poj3068

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题意：求两条从0~n-1的路径，使两条路径没有公共点并且变得权值的和最小，没有则输出Not possible

代码：

#include <vector>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;struct node{    int to,cap,cost,rev;};int V;int dis[10005],prevv[10005],preve[10005];vector<node> G[10005];void addedge(int from,int to,int cap,int cost){    G[from].push_back((node){to,cap,cost,G[to].size()});    G[to].push_back((node){from,0,-cost,G[from].size()-1});}int min_cost_flow(int s,int t,int f){    int i,d,v,ans,sign;    ans=0;    while(f>0){        fill(dis,dis+V,INF);        dis[s]=0;        sign=1;        while(sign){        sign=0;        for(v=0;v<V;v++){            if(dis[v]==INF)            continue;            for(i=0;i<G[v].size();i++){                node &e=G[v][i];                if(e.cap>0&&dis[e.to]>dis[v]+e.cost){                    dis[e.to]=dis[v]+e.cost;                    prevv[e.to]=v;                    preve[e.to]=i;                    sign=1;                }            }        }        }                                       //沿费用最短路增广        if(dis[t]==INF)        return -1;        d=f;        for(v=t;v!=s;v=prevv[v])        d=min(d,G[prevv[v]][preve[v]].cap);     //找出当前流量            f-=d;                   ans+=d*dis[t];        for(v=t;v!=s;v=prevv[v]){            node &e=G[prevv[v]][preve[v]];            e.cap-=d;            G[v][e.rev].cap+=d;                 //加反向边        }    }    return ans;}                                               //最小费用流模板   int main(){    int n,m,i,j,u,v,w,ans,cas;    cas=1;    while(scanf("%d%d",&n,&m)!=EOF&&(n||m)){        for(i=0;i<=n+1;i++)        G[i].clear();        addedge(0,1,2,0);                       //源点与第一个点容量为2，费用为0        addedge(n,n+1,2,0);                     //汇点与最后一个点容量也为2，费用为0        while(m--){            scanf("%d%d%d",&u,&v,&w);            addedge(u+1,v+1,1,w);               //建容量为1，权值为边权的边        }        V=n+2;        ans=min_cost_flow(0,V-1,2);        if(ans==-1)        printf("Instance #%d: Not possible\n",cas++);        else        printf("Instance #%d: %d\n",cas++,ans);    }    return 0;}