哎╮(╯▽╰)╭，这是费用流基础题型，拆点，建二分图，跑最小费用最大流即可。若最大流为n，则说明是最大匹配为n，所有点都参与，每个点的入度和出度又是1，所以就是环。

弱菜还需努力！

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int inf=0x3f3f3f3f;
int nume=0;int e[50000][4];int head[500];
int n,m;
void inline  adde(int i,int j,int c,int w)
{
    e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;
    e[nume][2]=c;e[nume++][3]=w;
    e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;
    e[nume][2]=0;e[nume++][3]=-w;
}
int inq[500];int pre[500];int prv[500];
int d[500];
bool spfa(int &sum,int &sumflow)
{
    for(int i=0;i<=2*n+2;i++)
          {
              inq[i]=0;
              d[i]=inf;
          }
    queue<int>q;
    q.push(0);
    inq[0]=1;
    d[0]=0;
    while(!q.empty())
    {
        int cur=q.front();
        q.pop();
        inq[cur]=0;
        for(int i=head[cur];i!=-1;i=e[i][1])
        {
            int v=e[i][0];
            if(e[i][2]>0&&d[cur]+e[i][3]<d[v])
            {
                d[v]=d[cur]+e[i][3];
                pre[v]=i;
                prv[v]=cur;
                if(!inq[v])
                {
                    q.push(v);
                    inq[v]=1;
                }
            }
        }
    }
    if(d[2*n+2]==inf)return 0;
    int cur=2*n+2;int minf=inf;
    while(cur!=0)
    {
        minf=e[pre[cur]][2]<minf?e[pre[cur]][2]:minf;
        cur=prv[cur];
    }
     cur=2*n+2;
    while(cur!=0)
    {
        e[pre[cur]][2]-=minf;
        e[pre[cur]^1][2]+=minf;
        cur=prv[cur];
    }
    sumflow+=minf;
    sum+=minf*d[2*n+2];
    return 1;
}

int mincost(int &sumflow)
{
    int sum=0;
    while(spfa(sum,sumflow));
    return sum;
}
void init()
{
    nume=0;
    memset(head,-1,sizeof(head));
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        int a,b,c;
        for(int i=0;i<m;i++)
        {
           scanf("%d%d%d",&a,&b,&c);
            adde(a,b+n,1,c);
        }
        for(int i=1;i<=n;i++)
        {
            adde(0,i,1,0);
            adde(i+n,2*n+1,1,0);
        }
        adde(2*n+1,2*n+2,n,0);
        int sumflow=0;
        int ans=mincost(sumflow);
        if(sumflow!=n)
        printf("-1\n");
        else
        {
            printf("%d\n",ans);
        }
    }
    return 0;
}