一道看起来比较纠结的题，但思路很好想：一天一天的递推。看开始没看到“The traveler always starts from city 1”，害得我想复杂了，差点儿把O（n^3）写成O（n^4）。

 1 #include<stdio.h>
 2 int m,n,f[105][105],cost[105][105];
 3 int main()
 4 {
 5     int i,j,k,max;
 6     while(scanf("%d%d",&n,&m)==2)
 7     {
 8         if(!m && !n) break;
 9         for(i = 1; i <= n; i++)
10             for(j = 1; j <= n; j++)
11                 scanf("%d",&cost[i][j]);
12         for(i = 1; i <= m; i++)
13             for(j = 1; j <= n; j++)
14                 scanf("%d",&f[i][j]);
15         f[0][1] = 0;
16         for(j = 1; j <= n; j++)
17             f[1][j] += f[0][1] - cost[1][j];
18         for(i = 2; i <= m; i++)
19             for(j = 1; j <= n; j++)
20             {
21                 max = -1<<20;
22                 for(k = 1; k <= n; k++)
23                 {
24                     if(f[i-1][k] - cost[k][j] > max)
25                         max = f[i-1][k] - cost[k][j];
26                 }
27                 f[i][j] += max;
28             }
29         max = 0;
30         for(j = 1; j <= n; j++)
31             if(f[m][j] > max) max = f[m][j];
32         printf("%d\n",max);
33     }
34     return 0;
35 }