最小费用，最大流的思想，也不知道怎么稀里糊涂就ac啦
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=102;
const int inf=2<<20;
int cost[maxn][maxn],flow[maxn][maxn],cap[maxn][maxn],d[maxn];
int n,m;
int solve()
{
queue<int> q;
int i,vis[maxn],p[maxn],tot=0;
while(1)
{
for(i=1;i<=n+1;i++)
{
d[i]=i==1?0:inf;
vis[i]=0;
}
q.push(1);
int x;
while(!q.empty())
{
x=q.front();
q.pop();
vis[x]=0;
for(i=1;i<=n+1;i++)
{
if(cap[x][i]>flow[x][i]&&d[i]>d[x]+cost[x][i])
{
d[i]=d[x]+cost[x][i];
p[i]=x;
if(!vis[i])
{
vis[i]=1;
q.push(i);
}
}
}
}
if(d[n+1]==inf) break;
int min=inf;
for(i=n+1;i!=1;i=p[i])
{
if(min>(cap[p[i]][i])-flow[p[i]][i])
min=cap[p[i]][i]-flow[p[i]][i];
}
for(i=n+1;i!=1;i=p[i])
{
flow[p[i]][i]+=min;
flow[i][p[i]]-=min;cost[i][p[i]]=-cost[i][p[i]];
}
tot+=min*d[n+1];
if(flow[n][n+1]==2) return tot;
}
return -1;
}
int main()
{
while(cin>>n&&n!=0)
{
cin>>m;
int i;
int tu,tv,tw;
memset(flow,0,sizeof(flow));
memset(cost,inf,sizeof(cost));
memset(cap,0,sizeof(cap));
for(i=1;i<=m;i++)
{
cin>>tu>>tv>>tw;
cost[tu][tv]=tw;cap[tu][tv]=1;
cost[tv][tu]=tw;cap[tv][tu]=1;
}
cap[n][n+1]=2;cost[n][n+1]=cost[n+1][n]=0;
int ou=solve();
if(ou!=-1) cout<<ou<<endl;
else cout<<"Back to jail"<<endl;
}
return 0;
}