【离散化 线段树 二分查找】3661可以被机器人摧毁的最大墙壁数目|分数未知

本文涉及知识点

【C++】树状数组的使用、原理、封装类、样例
C++线段树
C++二分查找

3661. 可以被机器人摧毁的最大墙壁数目

一条无限长的直线上分布着一些机器人和墙壁。给你整数数组 robots ，distance 和 walls：
robots[i] 是第 i 个机器人的位置。
distance[i] 是第 i 个机器人的子弹可以行进的 最大 距离。
walls[j] 是第 j 堵墙的位置。
每个机器人有 一颗 子弹，可以向左或向右发射，最远距离为 distance[i] 米。
子弹会摧毁其射程内路径上的每一堵墙。机器人是固定的障碍物：如果子弹在到达墙壁前击中另一个机器人，它会 立即 在该机器人处停止，无法继续前进。
返回机器人可以摧毁墙壁的 最大 数量。
注意：
墙壁和机器人可能在同一位置；该位置的墙壁可以被该位置的机器人摧毁。机器人不会被子弹摧毁。

示例 1:
输入: robots = [4], distance = [3], walls = [1,10]
输出: 1
解释:
robots[0] = 4 向 左 发射，distance[0] = 3，覆盖范围 [1, 4]，摧毁了 walls[0] = 1。
因此，答案是 1。
示例 2:
输入: robots = [10,2], distance = [5,1], walls = [5,2,7]

输出: 3

解释:
robots[0] = 10 向 左 发射，distance[0] = 5，覆盖范围 [5, 10]，摧毁了 walls[0] = 5 和 walls[2] = 7。
robots[1] = 2 向 左 发射，distance[1] = 1，覆盖范围 [1, 2]，摧毁了 walls[1] = 2。
因此，答案是 3。
示例 3:
输入: robots = [1,2], distance = [100,1], walls = [10]
输出: 0
解释:
在这个例子中，只有 robots[0] 能够到达墙壁，但它向 右 的射击被 robots[1] 挡住了，因此答案是 0。
提示:
1<=robots.length==distance.length<= 10 5 1 <= robots.length == distance.length <= 10^5 1<=robots.length==distance.length<=105
1<=walls.length<= 10 5 1 <= walls.length <= 10^5 1<=walls.length<=105
1<=robots[i],walls[j]<= 10 9 1 <= robots[i], walls[j] <= 10^9 1<=robots[i],walls[j]<=109
1<=distance[i]<= 10 5 1 <= distance[i] <= 10^5 1<=distance[i]<=105
robots 中的所有值都是 互不相同 的

错误解法 线段树+ 动态规划

动态规划的状态表示

dp[i]表示 只摧毁位置$\le i$的墙，最后能销毁多少堵墙。且之后不会消耗$\le i$的墙。
最大值线段树maxTree记录最大值。

动态规划的顺序

按机器人的位置从小到大处理。

动态规划的转移方程

每个机器人枚举两种状态，向左射击，向右射击。

动态规划的初始值

全为0.

动态规划的返回值

maxTree的最大值。

错误原因

由于两个机器人的射击范围不能重叠，否则会重复统计。故向左射击不一定是最大射程，各射程要一一枚举。
robots = { 4,10 }, distance = { 3,3 }, walls = { 6,7,8 };
第二个机器人向左射击能到{7,8,9,10}。第一个机器人向右射击到7，第二各机器人向左射击到8，才是正解。

错误代码

template<class TSave, class TRecord >class CRangUpdateLineTree{protected:virtual void OnQuery(TSave& ans,const TSave& save, const int& iSaveLeft, const int& iSaveRight) = 0;virtual void OnUpdate(TSave& save, const int& iSaveLeft, const int& iSaveRight, const TRecord& update) = 0;virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, const int& iSaveLeft, const int& iSaveRight) = 0;virtual void OnUpdateRecord(TRecord& old, const TRecord& newRecord) = 0;};template<class TSave, class TRecord >class CVectorRangeUpdateLineTree : public CRangUpdateLineTree<TSave, TRecord>{public:CVectorRangeUpdateLineTree(int iEleSize, TSave tDefault, TRecord tRecordNull) :m_iEleSize(iEleSize), m_tDefault(tDefault), m_save(iEleSize * 4, tDefault), m_record(iEleSize * 4, tRecordNull) {m_recordNull = tRecordNull;}void Update(int iLeftIndex, int iRightIndex, TRecord value){Update(1, 0, m_iEleSize - 1, iLeftIndex, iRightIndex, value);}TSave Query(int leftIndex, int rightIndex) {return Query(leftIndex, rightIndex, m_tDefault);}TSave Query(int leftIndex, int rightIndex,const TSave& tDefault) {TSave ans = tDefault;Query(ans, 1, 0, m_iEleSize - 1, leftIndex, rightIndex);return ans;}//void Init() {//Init(1, 0, m_iEleSize - 1);//}TSave QueryAll() {return m_save[1];}void swap(CVectorRangeUpdateLineTree<TSave, TRecord>& other) {m_save.swap(other.m_save);m_record.swap(other.m_record);std::swap(m_recordNull, other.m_recordNull);assert(m_iEleSize == other.m_iEleSize);}protected://void Init(int iNodeNO, int iSaveLeft, int iSaveRight)//{//if (iSaveLeft == iSaveRight) {//this->OnInit(m_save[iNodeNO], iSaveLeft);//return;//}//const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;//Init(iNodeNO * 2, iSaveLeft, mid);//Init(iNodeNO * 2 + 1, mid + 1, iSaveRight);//this->OnUpdateParent(m_save[iNodeNO], m_save[iNodeNO * 2], m_save[iNodeNO * 2 + 1], iSaveLeft, iSaveRight);//}void Query(TSave& ans,int iNodeNO, int iSaveLeft, int iSaveRight, int iQueryLeft, int iQueryRight) {if ((iSaveLeft >= iQueryLeft) && (iSaveRight <= iQueryRight)) {this->OnQuery(ans,m_save[iNodeNO], iSaveLeft, iSaveRight);return;}if (iSaveLeft == iSaveRight) {//没有子节点return;}Fresh(iNodeNO, iSaveLeft, iSaveRight);const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (mid >= iQueryLeft) {Query(ans,iNodeNO * 2, iSaveLeft, mid, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(ans,iNodeNO * 2 + 1, mid + 1, iSaveRight, iQueryLeft, iQueryRight);}}void Update(int iNode, int iSaveLeft, int iSaveRight, int iOpeLeft, int iOpeRight, TRecord value){if ((iOpeLeft <= iSaveLeft) && (iOpeRight >= iSaveRight)){this->OnUpdate(m_save[iNode], iSaveLeft, iSaveRight, value);this->OnUpdateRecord(m_record[iNode], value);return;}if (iSaveLeft == iSaveRight) {return;//没有子节点}Fresh(iNode, iSaveLeft, iSaveRight);const int iMid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (iMid >= iOpeLeft){Update(iNode * 2, iSaveLeft, iMid, iOpeLeft, iOpeRight, value);}if (iMid + 1 <= iOpeRight){Update(iNode * 2 + 1, iMid + 1, iSaveRight, iOpeLeft, iOpeRight, value);}// 如果有后代，至少两个后代this->OnUpdateParent(m_save[iNode], m_save[iNode * 2], m_save[iNode * 2 + 1], iSaveLeft, iSaveRight);}void Fresh(int iNode, int iDataLeft, int iDataRight){if (m_recordNull == m_record[iNode]){return;}const int iMid = iDataLeft + (iDataRight - iDataLeft) / 2;Update(iNode * 2, iDataLeft, iMid, iDataLeft, iMid, m_record[iNode]);Update(iNode * 2 + 1, iMid + 1, iDataRight, iMid + 1, iDataRight, m_record[iNode]);m_record[iNode] = m_recordNull;}vector<TSave> m_save;vector<TRecord> m_record;TRecord m_recordNull;TSave m_tDefault;const int m_iEleSize;};template<class TSave, class TRecord >class CTreeRangeLineTree : public CRangUpdateLineTree<TSave, TRecord>{protected:struct CTreeNode{int Cnt()const { return m_iMaxIndex - m_iMinIndex + 1; }int m_iMinIndex;int m_iMaxIndex;TRecord record;TSave data;CTreeNode* m_lChild = nullptr, * m_rChild = nullptr;};CTreeNode* m_root;TSave m_tDefault;TRecord m_tRecordDef;public:CTreeRangeLineTree(int iMinIndex, int iMaxIndex, TSave tDefault, TRecord tRecordDef) {m_tDefault = tDefault;m_tRecordDef = tRecordDef;m_root = CreateNode(iMinIndex, iMaxIndex);}void Update(int iLeftIndex, int iRightIndex, TRecord value){Update(m_root, iLeftIndex, iRightIndex, value);}TSave QueryAll() {return m_root->data;}TSave Query(int leftIndex, int leftRight) {TSave ans = m_tDefault;Query(ans,m_root, leftIndex, leftRight);return ans;}protected:void Query(TSave& ans, CTreeNode* node, int iQueryLeft, int iQueryRight) {if ((node->m_iMinIndex >= iQueryLeft) && (node->m_iMaxIndex <= iQueryRight)) {this->OnQuery(ans, node->data, node->m_iMinIndex, node->m_iMaxIndex);return;}if (1 == node->Cnt()) {//没有子节点return;}CreateChilds(node);Fresh(node);const int mid = node->m_iMinIndex + (node->m_iMaxIndex - node->m_iMinIndex) / 2;if (mid >= iQueryLeft) {Query(ans, node->m_lChild, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(ans, node->m_rChild, iQueryLeft, iQueryRight);}}void Update(CTreeNode* node, int iOpeLeft, int iOpeRight, TRecord value){const int& iSaveLeft = node->m_iMinIndex;const int& iSaveRight = node->m_iMaxIndex;if ((iOpeLeft <= iSaveLeft) && (iOpeRight >= iSaveRight)){this->OnUpdate(node->data, iSaveLeft, iSaveRight, value);this->OnUpdateRecord(node->record, value);return;}if (1 == node->Cnt()) {//没有子节点return;}CreateChilds(node);Fresh(node);const int mid = node->m_iMinIndex + (node->m_iMaxIndex - node->m_iMinIndex) / 2;if (mid >= iOpeLeft) {this->Update(node->m_lChild, iOpeLeft, iOpeRight, value);}if (mid + 1 <= iOpeRight) {this->Update(node->m_rChild, iOpeLeft, iOpeRight, value);}// 如果有后代，至少两个后代this->OnUpdateParent(node->data, node->m_lChild->data, node->m_rChild->data, node->m_iMinIndex, node->m_iMaxIndex);}void CreateChilds(CTreeNode* node) {if (nullptr != node->m_lChild) { return; }const int iSaveLeft = node->m_iMinIndex;const int iSaveRight = node->m_iMaxIndex;const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;node->m_lChild = CreateNode(iSaveLeft, mid);node->m_rChild = CreateNode(mid + 1, iSaveRight);}CTreeNode* CreateNode(int iMinIndex, int iMaxIndex) {CTreeNode* node = new CTreeNode;node->m_iMinIndex = iMinIndex;node->m_iMaxIndex = iMaxIndex;node->data = m_tDefault;node->record = m_tRecordDef;return node;}void Fresh(CTreeNode* node){if (m_tRecordDef == node->record){return;}CreateChilds(node);Update(node->m_lChild, node->m_lChild->m_iMinIndex, node->m_lChild->m_iMaxIndex, node->record);Update(node->m_rChild, node->m_rChild->m_iMinIndex, node->m_rChild->m_iMaxIndex, node->record);node->record = m_tRecordDef;}};template<class TSave, class TRecord >class CSetMaxLineTree : public CTreeRangeLineTree<TSave, TRecord>{public:using CTreeRangeLineTree<TSave, TRecord>::CTreeRangeLineTree;protected:virtual void OnQuery(TSave& ans, const TSave& save, const int& iSaveLeft, const int& iSaveRight) {ans = max(ans, save);}virtual void OnUpdate(TSave& save, const int& iSaveLeft, const int& iSaveRight, const TRecord& update) {save = update;}virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, const int& iSaveLeft, const int& iSaveRight) {par = max(left, r);}virtual void OnUpdateRecord(TRecord& old, const TRecord& newRecord){old = newRecord;}};class Solution {public:int maxWalls(vector<int>& robots, vector<int>& distance, vector<int>& walls) {const int N = robots.size();const int M = (int)(1e9 + 2e5);sort(walls.begin(), walls.end());vector<pair<int, int>> rd;for (int i = 0; i < N; i++) {rd.emplace_back(robots[i], distance[i]);}sort(rd.begin(), rd.end());vector<int> vLeft(N), vRight(N);for (int i = 0; i < N;i++) {const auto& [pos, dis] = rd[i];const int iLeftRobot = i ? rd[i - 1].first : 1;vLeft[i] = max(iLeftRobot, pos - dis);const int iRightPos = (i+1==N)? M : rd[i+1].first;vRight[i] = min(iRightPos, pos + dis);}CSetMaxLineTree<int, int> maxTree(0, M,0,0);for (int i = 0; i < N; i++) {const int x1 = max(1,vLeft[i]), x2 = rd[i].first, x3 = min(M,vRight[i]);const int cnt1 = upper_bound(walls.begin(), walls.end(), x2) - lower_bound(walls.begin(), walls.end(), x1);const int left = maxTree.Query(0, x1 - 1) + cnt1;const int cnt2 = upper_bound(walls.begin(), walls.end(), x3) - lower_bound(walls.begin(), walls.end(), x2);const int right = maxTree.Query(0, x2 - 1) + cnt2;maxTree.Update(x2,x2, left);maxTree.Update(x3, x3, right);}return maxTree.QueryAll();}}; 

正确解法

某个向右的机器人和某个向左的机器人射程重叠后。向右的机器人射程不变，向左的机器人缩短射程使之不重叠。向右射击仍然是一种状态，故只讨论向左。
令向左的机器人位于x2，向左能射击到x1。
则：射程非最大向左的最大值为 max ⁡x 1x 2 − 1 (dp[x]+f(x)) \\max_{x1}^{x2-1}( dp[x] + f(x)) maxx1x2−1​(dp[x]+f(x)),其中f(x)是处于$x+1\sim x2的墙数$,令g(x)是 $\le x$的墙数。
则 射程非最大向左的最大值为 max ⁡x 1x 2 − 1 (dp[x]+g(x2)−g(x))=g(x2)+ max ⁡x 1x 2 − 1 (dp[x]−g(x)) \\max_{x1}^{x2-1}( dp[x] +g(x2)-g(x))=g(x2)+\\max_{x1}^{x2-1}( dp[x] -g(x)) maxx1x2−1​(dp[x]+g(x2)−g(x))=g(x2)+maxx1x2−1​(dp[x]−g(x))
我们用最大值线段树maxTree2记录：dp[x]−g(x)
向左射击的最大值为：max(射程非最大向左的最大值,射程最大向左的最大值)

向右也要枚举

比如：

robots = {3,5 }, distance = { 2,2 }, walls = { 4,6 };

令向右的机器人在x2，向右能射击到x3。为了避免和前面的机器人重叠，我将此机器人的射程调整为：$x+1\to x3$
则起点非x2向右的最大值为： max ⁡x : x 2x 3 − 1 dp[x]+(x+1∼x3)的墙的个数=g(x3)+ max ⁡x : x 2x 3 − 1 (dp[x]−g(x)) \\max_{x:x2}^{x3-1}dp[x]+(x+1 \\sim x3)的墙的个数=g(x3)+\\max_{x:x2}^{x3-1}(dp[x]-g(x)) maxx:x2x3−1​dp[x]+(x+1∼x3)的墙的个数=g(x3)+maxx:x2x3−1​(dp[x]−g(x))
可以共用：maxTree2。

内存超限代码

template<class TSave, class TRecord >class CSingeUpdateLineTree{protected:virtual void OnQuery(TSave& ans,const TSave& cur) = 0;virtual void OnUpdate(TSave& save, int iSave, const TRecord& update) = 0;virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, int iSaveLeft, int iSaveRight) = 0;};template<class TSave, class TRecord >class CVectorSingUpdateLineTree : public CSingeUpdateLineTree<TSave, TRecord>{public:CVectorSingUpdateLineTree(int iEleSize, TSave tDefault) :m_iEleSize(iEleSize),m_save(iEleSize*4,tDefault), m_tDefault(tDefault){}void Update(int index, TRecord update) {Update(1, 0, m_iEleSize-1, index, update);}TSave Query(int leftIndex, int leftRight,TSave tDefault) {TSave ans = tDefault;Query(ans,1, 0, m_iEleSize - 1, leftIndex, leftRight);return ans;}TSave Query(int leftIndex, int leftRight) {return Query(leftIndex,leftRight, m_tDefault);}void Init(std::function<void(TSave&,const int&)> fun) {Init(fun,1, 0, m_iEleSize - 1);}TSave QueryAll() {return m_save[1];}protected:int m_iEleSize;void Init(std::function<void(TSave&, const int&)> fun,int iNodeNO, int iSaveLeft, int iSaveRight){if (iSaveLeft == iSaveRight) {fun(m_save[iNodeNO], iSaveLeft);return;}const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;Init(fun,iNodeNO * 2, iSaveLeft, mid);Init(fun,iNodeNO * 2 + 1, mid + 1, iSaveRight);this->OnUpdateParent(m_save[iNodeNO], m_save[iNodeNO*2], m_save[iNodeNO*2+1], iSaveLeft, iSaveRight);}void Query(TSave& ans,int iNodeNO, int iSaveLeft, int iSaveRight, int iQueryLeft, int iQueryRight) {if ((iSaveLeft >= iQueryLeft) && (iSaveRight <= iQueryRight)) {this->OnQuery(ans,m_save[iNodeNO]);return;}if (iSaveLeft == iSaveRight) {//没有子节点return;}const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (mid >= iQueryLeft) {Query(ans,iNodeNO * 2, iSaveLeft, mid, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(ans,iNodeNO * 2 + 1, mid + 1, iSaveRight, iQueryLeft, iQueryRight);}}void Update(int iNodeNO, int iSaveLeft, int iSaveRight, int iUpdateNO, TRecord update) {if (iSaveLeft == iSaveRight){this->OnUpdate(m_save[iNodeNO], iSaveLeft, update);return;}const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (iUpdateNO <= mid) {Update(iNodeNO * 2, iSaveLeft, mid, iUpdateNO, update);}else {Update(iNodeNO * 2 + 1, mid + 1, iSaveRight, iUpdateNO, update);}this->OnUpdateParent(m_save[iNodeNO], m_save[iNodeNO*2], m_save[iNodeNO*2+1], iSaveLeft, iSaveRight);}vector<TSave> m_save;const TSave m_tDefault;};template<class TSave, class TRecord >class CTreeSingeLineTree : public CSingeUpdateLineTree<TSave, TRecord>{protected:struct CTreeNode{int Cnt()const { return m_iMaxIndex - m_iMinIndex + 1; }int m_iMinIndex;int m_iMaxIndex;TSave data;CTreeNode* m_lChild=nullptr, *m_rChild=nullptr;~CTreeNode() {delete m_lChild; m_lChild = nullptr;delete m_rChild; m_rChild = nullptr;}};CTreeNode* m_root;TSave m_tDefault;public:CTreeSingeLineTree(int iMinIndex, int iMaxIndex, TSave tDefault) {m_tDefault = tDefault;m_root = CreateNode(iMinIndex, iMaxIndex);}void Update(int index, TRecord update) {Update(m_root, index, update);}TSave QueryAll() {return m_root->data;}TSave Query(int leftIndex, int leftRight) {TSave ans = m_tDefault;Query(ans,m_root, leftIndex, leftRight);return ans;}~CTreeSingeLineTree() {delete m_root;}protected:void Query(TSave& ans,CTreeNode* node, int iQueryLeft, int iQueryRight) {if ((node->m_iMinIndex >= iQueryLeft) && (node->m_iMaxIndex <= iQueryRight)) {this->OnQuery(ans,node->data);return;}if (1 == node->Cnt()) {//没有子节点return;}CreateChilds(node);const int mid = node->m_iMinIndex + (node->m_iMaxIndex - node->m_iMinIndex) / 2;if (mid >= iQueryLeft) {Query(ans,node->m_lChild, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(ans,node->m_rChild, iQueryLeft, iQueryRight);}}void Update(CTreeNode* node, int iUpdateNO, TRecord update) {if ((iUpdateNO < node->m_iMinIndex) || (iUpdateNO > node->m_iMaxIndex)) {return;}if (1 == node->Cnt()) {this->OnUpdate(node->data, node->m_iMinIndex, update);return;}CreateChilds(node);Update(node->m_lChild, iUpdateNO, update);Update(node->m_rChild, iUpdateNO, update);this->OnUpdateParent(node->data, node->m_lChild->data, node->m_rChild->data, node->m_iMinIndex, node->m_iMaxIndex);}void CreateChilds(CTreeNode* node) {if (nullptr != node->m_lChild) { return; }const int iSaveLeft = node->m_iMinIndex;const int iSaveRight = node->m_iMaxIndex;const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;node->m_lChild = CreateNode(iSaveLeft,mid);node->m_rChild = CreateNode(mid+1, iSaveRight);}CTreeNode* CreateNode(int iMinIndex, int iMaxIndex) {CTreeNode* node = new CTreeNode;node->m_iMinIndex = iMinIndex;node->m_iMaxIndex = iMaxIndex;node->data = m_tDefault;return node;}};template<class TSave, class TRecord >class CSetMaxLineTree : public CTreeSingeLineTree<TSave, TRecord>{public:using CTreeSingeLineTree<TSave, TRecord>::CTreeSingeLineTree;protected:virtual void OnQuery(TSave& ans, const TSave& cur) {ans = max(ans, cur);}virtual void OnUpdate(TSave& save, int iSave, const TRecord& updatee) {save = updatee;}virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, int iSaveLeft, int iSaveRight) {par = max(left, r);}};class Solution {public:int maxWalls(vector<int>& robots, vector<int>& distance, vector<int>& walls) {const int N = robots.size();const int M = (int)(1e9 + 2e5);sort(walls.begin(), walls.end());vector<pair<int, int>> rd;for (int i = 0; i < N; i++) {rd.emplace_back(robots[i], distance[i]);}sort(rd.begin(), rd.end());vector<int> vLeft(N), vRight(N);for (int i = 0; i < N;i++) {const auto& [pos, dis] = rd[i];const int iLeftRobot = i ? rd[i - 1].first : 1;vLeft[i] = max(iLeftRobot, pos - dis);const int iRightPos = (i+1==N)? M : rd[i+1].first;vRight[i] = min(iRightPos, pos + dis);}CSetMaxLineTree<int, int> maxTree(0, M,0), maxTree2(0, M, -1000\'000);for (int i = 0; i < N; i++) {const int x1 = max(1,vLeft[i]), x2 = rd[i].first, x3 = min(M,vRight[i]);const int g2 = upper_bound(walls.begin(), walls.end(), x2) - walls.begin();const int g3 = upper_bound(walls.begin(), walls.end(), x3) - walls.begin();const int cnt1 = upper_bound(walls.begin(), walls.end(), x2) - lower_bound(walls.begin(), walls.end(), x1);const int left = maxTree.Query(0, x1 - 1) + cnt1;const int cnt2 = upper_bound(walls.begin(), walls.end(), x3) - lower_bound(walls.begin(), walls.end(), x2);const int right = maxTree.Query(0, x2 - 1) + cnt2;const int left2 = maxTree2.Query(x1,x2-1)+g2;const int right2 = maxTree2.Query(x2, x3 - 1) + g3;maxTree.Update(x2, max(left,left2));maxTree.Update(x3, max(right,right2));maxTree2.Update(x2, max(left, left2) -g2);maxTree2.Update(x3, max(right, right2) -g3);}return maxTree.QueryAll();}}; 

空间超限的解决方法

一，离散化。
二，改用最大值树状数组。

核心代码

template<class T=int>class CDiscretize //离散化{public:CDiscretize(vector<T> nums){sort(nums.begin(), nums.end());nums.erase(std::unique(nums.begin(), nums.end()), nums.end());m_nums = nums;for (int i = 0; i < nums.size(); i++){m_mValueToIndex[nums[i]] = i;}}int operator[](const T value)const{auto it = m_mValueToIndex.find(value);if (m_mValueToIndex.end() == it){return -1;}return it->second;}int size()const{return m_mValueToIndex.size();}vector<T> m_nums;protected:unordered_map<T, int> m_mValueToIndex;};template<class TSave, class TRecord >class CSingeUpdateLineTree{protected:virtual void OnQuery(TSave& ans,const TSave& cur) = 0;virtual void OnUpdate(TSave& save, int iSave, const TRecord& update) = 0;virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, int iSaveLeft, int iSaveRight) = 0;};template<class TSave, class TRecord >class CVectorSingUpdateLineTree : public CSingeUpdateLineTree<TSave, TRecord>{public:CVectorSingUpdateLineTree(int iEleSize, TSave tDefault) :m_iEleSize(iEleSize),m_save(iEleSize*4,tDefault), m_tDefault(tDefault){}void Update(int index, TRecord update) {Update(1, 0, m_iEleSize-1, index, update);}TSave Query(int leftIndex, int leftRight,TSave tDefault) {TSave ans = tDefault;Query(ans,1, 0, m_iEleSize - 1, leftIndex, leftRight);return ans;}TSave Query(int leftIndex, int leftRight) {return Query(leftIndex,leftRight, m_tDefault);}void Init(std::function<void(TSave&,const int&)> fun) {Init(fun,1, 0, m_iEleSize - 1);}TSave QueryAll() {return m_save[1];}protected:int m_iEleSize;void Init(std::function<void(TSave&, const int&)> fun,int iNodeNO, int iSaveLeft, int iSaveRight){if (iSaveLeft == iSaveRight) {fun(m_save[iNodeNO], iSaveLeft);return;}const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;Init(fun,iNodeNO * 2, iSaveLeft, mid);Init(fun,iNodeNO * 2 + 1, mid + 1, iSaveRight);this->OnUpdateParent(m_save[iNodeNO], m_save[iNodeNO*2], m_save[iNodeNO*2+1], iSaveLeft, iSaveRight);}void Query(TSave& ans,int iNodeNO, int iSaveLeft, int iSaveRight, int iQueryLeft, int iQueryRight) {if ((iSaveLeft >= iQueryLeft) && (iSaveRight <= iQueryRight)) {this->OnQuery(ans,m_save[iNodeNO]);return;}if (iSaveLeft == iSaveRight) {//没有子节点return;}const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (mid >= iQueryLeft) {Query(ans,iNodeNO * 2, iSaveLeft, mid, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(ans,iNodeNO * 2 + 1, mid + 1, iSaveRight, iQueryLeft, iQueryRight);}}void Update(int iNodeNO, int iSaveLeft, int iSaveRight, int iUpdateNO, TRecord update) {if (iSaveLeft == iSaveRight){this->OnUpdate(m_save[iNodeNO], iSaveLeft, update);return;}const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (iUpdateNO <= mid) {Update(iNodeNO * 2, iSaveLeft, mid, iUpdateNO, update);}else {Update(iNodeNO * 2 + 1, mid + 1, iSaveRight, iUpdateNO, update);}this->OnUpdateParent(m_save[iNodeNO], m_save[iNodeNO*2], m_save[iNodeNO*2+1], iSaveLeft, iSaveRight);}vector<TSave> m_save;const TSave m_tDefault;};template<class TSave, class TRecord >class CTreeSingeLineTree : public CSingeUpdateLineTree<TSave, TRecord>{protected:struct CTreeNode{int Cnt()const { return m_iMaxIndex - m_iMinIndex + 1; }int m_iMinIndex;int m_iMaxIndex;TSave data;CTreeNode* m_lChild=nullptr, *m_rChild=nullptr;~CTreeNode() {delete m_lChild; m_lChild = nullptr;delete m_rChild; m_rChild = nullptr;}};CTreeNode* m_root;TSave m_tDefault;public:CTreeSingeLineTree(int iMinIndex, int iMaxIndex, TSave tDefault) {m_tDefault = tDefault;m_root = CreateNode(iMinIndex, iMaxIndex);}void Update(int index, TRecord update) {Update(m_root, index, update);}TSave QueryAll() {return m_root->data;}TSave Query(int leftIndex, int leftRight) {TSave ans = m_tDefault;Query(ans,m_root, leftIndex, leftRight);return ans;}~CTreeSingeLineTree() {delete m_root;}protected:void Query(TSave& ans,CTreeNode* node, int iQueryLeft, int iQueryRight) {if ((node->m_iMinIndex >= iQueryLeft) && (node->m_iMaxIndex <= iQueryRight)) {this->OnQuery(ans,node->data);return;}if (1 == node->Cnt()) {//没有子节点return;}CreateChilds(node);const int mid = node->m_iMinIndex + (node->m_iMaxIndex - node->m_iMinIndex) / 2;if (mid >= iQueryLeft) {Query(ans,node->m_lChild, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(ans,node->m_rChild, iQueryLeft, iQueryRight);}}void Update(CTreeNode* node, int iUpdateNO, TRecord update) {if ((iUpdateNO < node->m_iMinIndex) || (iUpdateNO > node->m_iMaxIndex)) {return;}if (1 == node->Cnt()) {this->OnUpdate(node->data, node->m_iMinIndex, update);return;}CreateChilds(node);Update(node->m_lChild, iUpdateNO, update);Update(node->m_rChild, iUpdateNO, update);this->OnUpdateParent(node->data, node->m_lChild->data, node->m_rChild->data, node->m_iMinIndex, node->m_iMaxIndex);}void CreateChilds(CTreeNode* node) {if (nullptr != node->m_lChild) { return; }const int iSaveLeft = node->m_iMinIndex;const int iSaveRight = node->m_iMaxIndex;const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;node->m_lChild = CreateNode(iSaveLeft,mid);node->m_rChild = CreateNode(mid+1, iSaveRight);}CTreeNode* CreateNode(int iMinIndex, int iMaxIndex) {CTreeNode* node = new CTreeNode;node->m_iMinIndex = iMinIndex;node->m_iMaxIndex = iMaxIndex;node->data = m_tDefault;return node;}};template<class TSave, class TRecord >class CSetMaxLineTree : public CVectorSingUpdateLineTree<TSave, TRecord>{public:using CVectorSingUpdateLineTree<TSave, TRecord>::CVectorSingUpdateLineTree;protected:virtual void OnQuery(TSave& ans, const TSave& cur) {ans = max(ans, cur);}virtual void OnUpdate(TSave& save, int iSave, const TRecord& updatee) {save = max(save,updatee);}virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, int iSaveLeft, int iSaveRight) {par = max(left, r);}};class Solution {public:int maxWalls(vector<int>& robots, vector<int>& distance, vector<int>& walls) {N = robots.size();Init(robots,distance,walls);CSetMaxLineTree<int, int> maxTree(M+1,0), maxTree2(M+1, -1000\'000);for (const auto&[x1,x2,x3]: m_xs) {const int g2 = upper_bound(walls.begin(), walls.end(), x2) - walls.begin();const int g3 = upper_bound(walls.begin(), walls.end(), x3) - walls.begin();const int cnt1 = upper_bound(walls.begin(), walls.end(), x2) - lower_bound(walls.begin(), walls.end(), x1);const int left = maxTree.Query(0, x1 - 1) + cnt1;const int cnt2 = upper_bound(walls.begin(), walls.end(), x3) - lower_bound(walls.begin(), walls.end(), x2);const int right = maxTree.Query(0, x2 - 1) + cnt2;const int left2 = maxTree2.Query(x1,x2-1)+g2;const int right2 = maxTree2.Query(x2, x3 - 1) + g3;maxTree.Update(x2, max(left,left2));maxTree.Update(x3, max(right,right2));maxTree2.Update(x2, max(left, left2) -g2);maxTree2.Update(x3, max(right, right2) -g3);}return maxTree.QueryAll();}void Init(const vector<int>& robots, const vector<int>& distance, vector<int>& walls) {vector<pair<int, int>> rd;sort(walls.begin(), walls.end());auto tmp = walls;tmp.emplace_back(INT_MIN / 2);//编码增加0,实际编码从1开始for (int i = 0; i < N; i++) {rd.emplace_back(robots[i], distance[i]);tmp.emplace_back(robots[i]);}CDiscretize<int> disc(tmp);for (auto& i : walls) {i = disc[i];}sort(rd.begin(), rd.end());for (int i = 0; i < N; i++) {const auto& [pos, dis] = rd[i];const int iLeftRobot = i ? rd[i - 1].first : 1;const int iLeft = max(iLeftRobot, pos - dis);const int iRightRobot = (i + 1 == N) ? (INT_MAX/2) : rd[i + 1].first;const int iRight = min(iRightRobot, pos + dis);const int x1 = lower_bound(disc.m_nums.begin(), disc.m_nums.end(), iLeft) - disc.m_nums.begin();const int x2 = disc[pos];const int x3 = upper_bound(disc.m_nums.begin(), disc.m_nums.end(), iRight) - disc.m_nums.begin()-1;m_xs.emplace_back(x1, x2, x3);}M = disc.m_nums.size();}int N, M;vector<tuple<int, int, int>> m_xs;}; 

单元测试

vector<int> robots, distance, walls;TEST_METHOD(TestMethod00){robots = { 4,10 }, distance = { 3,3 }, walls = { 6,7,8 };auto res = Solution().maxWalls(robots, distance, walls);AssertEx(3, res);}TEST_METHOD(TestMethod01){robots = {3,5 }, distance = { 2,2 }, walls = { 4,6 };auto res = Solution().maxWalls(robots, distance, walls);AssertEx(2, res);}TEST_METHOD(TestMethod11){robots = { 4 }, distance = { 3 }, walls = { 1,10 };auto res = Solution().maxWalls(robots, distance, walls);AssertEx(1, res);}TEST_METHOD(TestMethod12){robots = { 10,2 }, distance = { 5,1 }, walls = { 5,2,7 };auto res = Solution().maxWalls(robots, distance, walls);AssertEx(3, res);}TEST_METHOD(TestMethod13){robots = { 1,2 }, distance = { 100,1 }, walls = { 10 };auto res = Solution().maxWalls(robots, distance, walls);AssertEx(0, res);}TEST_METHOD(TestMethod14){robots = { 17, 59, 32, 11, 72, 18 }, distance = { 5, 7, 6, 5, 2, 10 },walls = {17, 25, 33, 29, 54, 53, 18, 35, 39, 37, 20, 14, 34, 13, 16, 58, 22, 51, 56, 27, 10, 15, 12, 23, 45, 43, 21, 2, 42, 7, 32, 40, 8, 9, 1, 5, 55, 30, 38, 4, 3, 31, 36, 41, 57, 28, 11, 49, 26, 19, 50, 52, 6, 47, 46, 44, 24, 48};auto res = Solution().maxWalls(robots, distance, walls);AssertEx(37, res);}TEST_METHOD(TestMethod15){robots = { 31,22,4,43,8,38,5,15,35,37,27,42,40,28,20,21 }, distance = { 3,5,5,7,8,1,10,7,9,6,3,4,4,5,7,4 },walls = { 34,74,54,46,79,89,7,73,12,27,44,5,62,43,60,71,10,63,41,77,33,91,32,53,66,51,78,18,61,6,8,24,23,81,3,25,40,85,84,15,52,48,17,59,55,64,50,21,88,36,2,16,80,69,22,87,1,28,65,31,83,26,67,72,29,75,57,9,30,86,39,37,13,19,56,68,35,90 };auto res = Solution().maxWalls(robots, distance, walls);AssertEx(41, res);}

简单版

性质一：如果机器人和墙挨着一起，此墙一定被摧毁。故只考虑不挨着机器人的墙。
性质二：由于子弹遇到机器人会停止，故墙只会被相邻的机器人摧毁。比赛的时候，没有挖掘到此性质。

动态规划的状态表示

dp0n表示第$0\sim i$号机器人都已经射击完毕，且最后一个机器人向左（右)射击能摧毁最后的墙数。
空间复杂度：O(n)
为了方便处理边界情况，增加一个机器人标兵，射击距离都是0，位置分别在正负无穷大。

动态规划的填表顺序

n = 1 to N 枚举后继状态和选择。

动态规划的转移方程

如果 i-1 和 i 都向左射击，两者不会摧毁同一道墙。
如果 i-1 向右和 i 向左射击，两者可能摧毁同一道墙。需要出度重复。i向右能够摧毁g1道墙，i-1向左能够摧毁g0道墙，i和i-1之间的机器人数量c，则重复的墙数为：$max(0,g1+g0-c)$。
如果i向右射击，i-1无论向左还是向右，两者都不会摧毁同一道墙。

动态规划的初始值

全为0。

动态规划的返回值

$\max (dp0.back(),dp1.back())$

代码

class Solution {public:int maxWalls(vector<int>& robots, vector<int>& distance, vector<int>& walls) {const int N = robots.size();const int M = int(1e9 + 1e5 + 1);sort(walls.begin(), walls.end());vector<pair<int, int>> rd;for (int i = 0; i < N; i++) {rd.emplace_back(robots[i], distance[i]);}rd.emplace_back(INT_MIN / 2, 0);rd.emplace_back(INT_MAX / 2, 0);sort(rd.begin(), rd.end());vector<int> vLeft(N+2), vRight(N+2);for (int i = 0; i < rd.size(); i++) {const auto& [pos, dis] = rd[i];const int iLeftRobot = i ? rd[i - 1].first : -M;vLeft[i] = max(iLeftRobot, pos - dis-1);const int iRightRobot = (i + 1 == N+2) ? M : rd[i + 1].first;vRight[i] = min(iRightRobot, pos + dis+1);}auto Cnt = [&](int left, int r) {int ans = lower_bound(walls.begin(), walls.end(), r)- upper_bound(walls.begin(), walls.end(), left);return ans;};//（left,r)之间的墙数量，不包括left,r。vector<int> dp0(N + 2), dp1(N + 2);for (int n = 1; n <= N; n++) {dp1[n] = max(dp0[n - 1], dp1[n - 1]) + Cnt(rd[n].first,vRight[n]);const int g = Cnt(vLeft[n], rd[n].first);dp0[n] = dp0[n-1] + g;const int iRepeat = g + Cnt(rd[n-1].first,vRight[n-1])- Cnt(rd[n - 1].first, rd[n].first);dp0[n] = max(dp0[n], dp1[n - 1] + g - max(0,iRepeat));}int cntSamePos = 0;for (const auto& pos : robots) {cntSamePos += Cnt(pos-1, pos+1);}return cntSamePos+max(dp0[N], dp1[N]);}};

扩展阅读

我想对大家说的话 工作中遇到的问题，可以按类别查阅鄙人的算法文章，请点击《算法与数据汇总》。 学习算法：按章节学习《喜缺全书算法册》，大量的题目和测试用例，打包下载。重视操作 有效学习：明确的目标 及时的反馈 拉伸区（难度合适） 专注 员工说：技术至上，老板不信；投资人的代表说：技术至上，老板会信。 闻缺陷则喜(喜缺)是一个美好的愿望，早发现问题，早修改问题，给老板节约钱。 子墨子言之：事无终始，无务多业。也就是我们常说的专业的人做专业的事。 如果程序是一条龙，那算法就是他的是睛 失败+反思=成功 成功+反思=成功

视频课程

先学简单的课程，请移步CSDN学院，听白银讲师（也就是鄙人）的讲解。
https://edu.csdn.net/course/detail/38771
如何你想快速形成战斗了，为老板分忧，请学习C#入职培训、C++入职培训等课程
https://edu.csdn.net/lecturer/6176

测试环境

操作系统：win7 开发环境： VS2019 C++17
或者 操作系统：win10 开发环境： VS2022 C++17
如无特殊说明，本算法用**C++**实现。

服装新闻