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
");
else
{
printf("%d
",ans);
}
}
return 0;
}