Floyd求最小环！（转载，非原创）

//Floyd 的 改进写法可以解决最小环问题，时间复杂度依然是 O(n^3)，储存结构也是邻接矩阵

int mincircle = infinity;
Dist = Graph;

for(int k=0;k<nVertex;++k){
    //新增部分:
    for(int i=0;i<k;++i)
        for(int j=0;j<i;++j)
             mincircle = min(mincircle,Dist[i][j]+Graph[j][k]+Graph[k][i]);
    //通常的 floyd 部分:
    for(int i=0;i<nVertex;++i)
        for(int j=0;j<i;++j){
            int temp = Dist[i][k] + Disk[k][j];
            if(temp < Dist[i][j])
                 Dist[i][j] = Dist[j][i] = temp;
         }
}

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
#define INF 0x3f3f3f3f
int n,m;
int map_[1000][1000];
int dis[1000][1000];
int main()
{
    while (cin>>n>>m)
    {
        memset(map_,INF,sizeof(map_));
        for (int i=1; i<=m; i++)
        {
            int a,b,c;
            cin>>a>>b>>c;
            //cout<<a<<b<<c<<endl;
            map_[a][b]=map_[b][a]=min(c,map_[a][b]);
           // cout<<map_[a][b]<<map_[b][a]<<endl;
        }
        for (int i=1; i<=n; i++)
        {
            for (int j=1; j<=n; j++)
            {
               // cout<<map_[i][j]<<" ";
                dis[i][j]=map_[i][j];
            }//cout<<endl;;
        }
        int ans=INF;
        for (int k=1; k<=n; k++)
        {
            for (int i=1; i<=n; i++)
            {
                for(int j=1; j<=n; j++)
                {
                    if (i!=j&&j!=k&&i!=k&&map_[i][k]!=INF&&map_[k][j]!=INF&&dis[i][j]!=INF)
                    {
                        ans=min(ans,dis[i][j]+map_[j][k]+map_[k][i]);
                        //cout<<ans<<endl;
                    }
                }
            }
            for (int i=1; i<=n; i++)
            {
                for (int j=1; j<=n; j++)
                {
                    dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
                }
            }
        }
        if (ans==INF) cout<<"It‘s impossible."<<endl;
        else
            cout<<ans<<endl;
    }
    return 0;
}