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題意：給定v個點和e條邊的有向加權圖，求1~v的兩條不相交(除了終點和起點沒有公共點)的路徑，使得路徑最小。

題解：把2到v-1的每個結點拆成i和i‘兩個結點，中間連一條容量為1，費用為0的邊，然後求1到v的流量為2的最小費用流即可。

附上程式碼：

#include<bits/stdc++.h>

using namespace std;

const int maxn=2e3+50;
const int inf=0x3f3f3f3f;

struct Edge{
    int from,to,cap,flow,cost;
    Edge(int _from,int _to,int _cap,int _flow,int _cost):from(_from),to(_to),cap(_cap),flow(_flow),cost(_cost){}
};

struct MCMF{
    int n,m;
    vector<Edge>edges;
    vector<int>G[maxn];
    int inq[maxn];
    int d[maxn];
    int p[maxn];
    int a[maxn];

    void init(int n){
        this->n=n;
        for(int i=0;i<n;i++){
            G[i].clear();
        }
        edges.clear();
    }

    void addedge(int from,int to,int cap,int cost){
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool bellmanford(int s,int t,int flow_limit,int &flow,int &cost){
        for(int i=0;i<n;i++){
            d[i]=inf;
        }
        memset(inq,0,sizeof(inq));
        d[s]=0;inq[s]=1;
        p[s]=0;a[s]=inf;
        queue<int>Q;
        Q.push(s);
        while(!Q.empty()){
            int u=Q.front();Q.pop();
            inq[u]=0;
            for(int i=0;i<G[u].size();i++){
                Edge&e=edges[G[u][i]];
                if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
                    d[e.to]=d[u]+e.cost;
                    p[e.to]=G[u][i];
                    a[e.to]=min(a[u],e.cap-e.flow);
                    if(!inq[e.to]){
                        Q.push(e.to);
                        inq[e.to]=1;
                    }
                }
            }
        }
        if(d[t]==inf){
            return false;
        }
        if(flow+a[t]>flow_limit){
            a[t]=flow_limit-flow;
        }
        flow+=a[t];
        cost+=d[t]*a[t];
        for(int u=t;u!=s;u=edges[p[u]].from){
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
        }
        return true;
    }

    int mincostflow(int s,int t,int flow_limit,int &cost){
        int flow=0;
        cost=0;
        while(flow<flow_limit&&bellmanford(s,t,flow_limit,flow,cost));
        return flow;
    }
};

MCMF g;

int main()
{
    int n,m,a,b,c;
    while(scanf("%d%d",&n,&m)==2&&n){
        g.init(n*2-2);
        for(int i=2;i<=n-1;i++){
            g.addedge(i-1,i+n-2,1,0);
        }
        while(m--){
            scanf("%d%d%d",&a,&b,&c);
            if(a!=1&&a!=n){
                a+=n-2;
            }else{
                a--;
            }
            b--;
            g.addedge(a,b,1,c);
        }
        int cost;
        g.mincostflow(0,n-1,2,cost);
        printf("%d\n",cost);
    }
    return 0;
}