最短Hamilton路径 数位dp

最短Hamilton路径

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 21;
 4 int dp[1<<maxn][maxn];
 5 int maps[maxn][maxn];
 6 int main() {
 7     int n; cin >> n;
 8     for (int i = 0; i < n; i++)
 9         for (int j = 0; j < n; j++)
10             cin >> maps[i][j];
11     memset(dp,0x3f3f3f3f,sizeof(dp));
12     dp[0][0] = 0;
13     for (int i = 0; i < (1<<n); i++) {
14         for (int j = 0; j < n; j++) {
15             if ((i>>j)&1) {
16                 for (int k = 0; k < n; k++) {
17                     if ((i>>k)&1)
18                         dp[i][j] = min(dp[i][j],dp[i^(1<<j)][k]+maps[k][j]);
19                 }
20             }
21         }
22     }
23     printf("%d\n",dp[(1<<n)-1][n-1]);
24     return 0;
25 }

原文地址：https://www.cnblogs.com/wstong/p/11764104.html