按層dp，f[i][j]表示已擴展i子集的節點當前在第j層的最小代價，預處理點集間距離即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 12
#define inf 1000000000
char getc(){char c=getchar();while ((c<‘
 
A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1
 
)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N][N],f[1<<N][N],dis[1<<N][1<<N];
int main()
{
    n=read(),m=read();
    memset(a,42,sizeof(a));
    for (int i=1;i<=m;i++)
    {
        int x=read()-1,y=read()-1,z=read();
        a[x][y]=min(a[x][y],z),a[y][x]=min(a[y][x],z);
    }
    memset(dis,
 
42,sizeof(dis));
    for (int i=1;i<(1<<n);i++)
    if ((i&-i)<i)
    {
        for (int j=0;j<n;j++)
        if (i&(1<<j))
            for (int k=0;k<n;k++)
            if ((i&(1<<k))&&j!=k) dis[i][1<<j]=min(dis[i][1<<j],a[j][k]);
        for (int j=i;j;j=j-1&i)
        if ((j&-j)<j) dis[i][j]=min(dis[i^j^(j&-j)][j&-j]+dis[i^(j&-j)][j^(j&-j)],inf);
    }
    memset(f,42,sizeof(f));
    for (int i=0;i<n;i++) f[1<<i][1]=0;
    for (int k=2;k<=n;k++)
        for (int i=1;i<(1<<n);i++)
            for (int j=i;j;j=j-1&i)
            if (dis[i][j]<100000000) f[i][k]=min(f[i][k],f[i^j][k-1]+dis[i][j]*(k-1));
    int ans=inf;
    for (int i=1;i<=n;i++) ans=min(ans,f[(1<<n)-1][i]);
    cout<<ans;
    return 0;
}

Luogu3959 NOIP2017寶藏（狀壓dp）