dp[i][j]:状态为i,当前到达节点是j的最短路径
状态转移方程:dp[(1<< j)|i][j] = Min(dp[(1<< j)|i][j], dp[i][k]+a[k][j]);
宏定义的min比stl快
#include <cstdio>
#include <cstring>
#include <algorithm>
#define Min(a, b) a>b?b:a
using namespace std;
int n, a[20][20], dp[1<<20][20];
int main()
{
scanf("%d", &n);
int i, j, k;
for(i = 0; i < n; ++i)
for(j = 0; j < n; ++j)
scanf("%d", &a[i][j]);
memset(dp, 0x3f, sizeof(dp));
dp[1][0] = 0;
for(i = 0; i < (1<<n); ++i)//枚举状态
for(j = 0; j < n; ++j)//枚举将要到达的地点
{
if((1<<j)&i) continue;
for(k = 0; k < n; ++k)//枚举中间节点
{
if(!((1<<k)&i)) continue;
dp[(1<<j)|i][j] = Min(dp[(1<<j)|i][j], dp[i][k]+a[k][j]);
}
}
int ans = 0x3f3f3f3f;
for(i = 1; i < n; ++i)
ans = Min(ans, dp[(1<<n)-1][i]+a[i][0]);
printf("%d\n", ans);
return 0;
}