poj3230Travel(dp动态规划)

题目链接：http://poj.org/problem?id=3230

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<cstdio>
 6 #define inf 0x3f3f3f3f
 7 using namespace std;
 8 const int N = 105;
 9 
10 int dp[N][N], in[N][N], to[N][N];
11 
12 int main() {
13 
14     int n, m;
15     while (cin >> n >> m && n != 0 && m != 0) {
16         memset(dp, -0x3f, sizeof(dp));
17         for (int i = 1; i <= n; i++) {
18             //dp[0][i] = 0;
19             for (int j = 1; j <= n; j++) {
20                 cin >> to[i][j];
21             }
22         }
23         for (int i = 1; i <= m; i++) {
24             for (int j = 1; j <= n; j++) {
25                 cin >> in[i][j];
26                 if (i == 1) {
27                     dp[i][j] = in[i][j] - to[i][j];
28                 }
29             }
30         }
31         //cout << dp[1][1] << dp[1][2] << dp[1][3] << "\n";
32         for (int i = 2; i <= m; i++) {
33             for (int j = 1; j <= n; j++) {
34                 for (int k = 1; k <= n; k++) {
35                     dp[i][j] = max(dp[i - 1][k] + in[i][j] - to[k][j], dp[i][j]);
36                     //cout << dp[i][j] << " " << i << " " << j << "\n";
37                 }
38             }
39         }
40         int ans = -inf;
41         for (int i = 1; i <= n; i++) {
42             ans = max(dp[m][i], ans);
43         }
44         cout << ans << "\n";
45     }
46     return 0;
47 }

 

posted @ 2022-03-25 19:33  Keyzee  阅读(27)  评论(0)    收藏  举报