NOIP 2009 普及组 道路游戏

• 时间复杂度O (n3 O(n^3 O(n3)

#include using namespace std;const int N = 1009;int n, m, p;int gold[N][N];//gold[i][j]:i号条路j时刻出现的金币int cost[N];//cost[i]:i号工厂购买机器人的费用 int f[N];//f[j]:j时刻得到的最多金币 int main() {cin >> n >> m >> p;for (int i = 1; i <= n; i ++) {for (int j = 1; j <= m; j ++) {cin >> gold[i][j];}}for (int i = 1; i <= n; i ++) cin >> cost[i];for (int j = 1; j <= m; j ++) {for (int i = 1; i <= n; i ++) {//j-1时刻在i-1号马路机器人停止行走 //j时刻在i号马路购买一个机器人，行走k次int tmp = f[j - 1] - cost[i]; for (int k = 1; k <= min(m - j + 1, p); k ++) {int road = (i - 1) + k;if (road > n) road -= n;int time = (j - 1) + k;tmp += gold[road][time];f[time] = max(f[time], tmp);}}}cout << f[m];return 0;}