思路:

状压dp。

实现:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring> 
 4 using namespace std;
 5 #define N 16
 6 #define INF 0x3f3f3f3f
 7 int map[N][N];
 8 int dp[1 << N][N];
 9 int rec(int S, int v, int n)
10 {
11     if(dp[S][v] >= 0)
12         return dp[S][v];
13     if(S == (1 << n) - 1 && v == n - 1)
14     {
15         return dp[S][v] = 0;
16     }
17     int res = INF;
18     for(int i = 0; i < n; i++)
19     {
20         if(!(S >> i & 1))
21         {
22             res = min(res, rec(S | 1 << i, i, n) + map[v][i]);
23         }
24     }
25     return dp[S][v] = res;
26 }
27 int main()
28 {
29     memset(dp, -1, sizeof(dp));
30     int n;
31     scanf("%d", &n);
32     for(int i = 0; i < n; i++)
33     {
34         for(int j = 0; j < n; j++)
35         {
36             scanf("%d",&map[i][j]);
37             if(j == 0 && i != 0)
38             {
39                 map[i][j] = INF;
40             }
41         }
42     }
43     cout << rec(0, 0, n) << endl;
44     return 0;
45 }

 

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