poj 3068 "Shortest" pair of paths

题目连接：“Shortest” pair of paths

题目大意：给你一个有向图，有边权，现在需要你找出两条从0到n-1的路径并且两条路径无公共点而且需要使得边权和最小，不存在输出No possible，存在输出两条路径的边权和

题目思路：这题我们可以想到去跑网络流，然后看最大流是不是2，我只给2的流量从超级源点流出，为了方便起见，我们将所有点位置加1，变成求1到n的最小费用最大流，超级源点为0，连接1，容量为2，单位费用为0，然后图中给的边容量为1（因为不能有公共点），费用为题目给的边权，然后第n个点连接超级汇点n+1,容量为2，费用为0，然后跑最小费用最大流就好了，判断满流以确定是不是有这样的两条路径

#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int MaxNode = 205;const int MaxEdge = 40005;struct Edge{    int to,vol,cost,next;}Edges[MaxEdge];int Pre[MaxNode],Path[MaxNode],Dist[MaxNode],Head[MaxNode],EdgeCount;void addedge(int u, int v, int vol, int cost){    Edges[EdgeCount].to = v;    Edges[EdgeCount].vol = vol;    Edges[EdgeCount].cost = cost;    Edges[EdgeCount].next = Head[u];    Head[u] = EdgeCount++;    Edges[EdgeCount].to = u;    Edges[EdgeCount].vol = 0;    Edges[EdgeCount].cost = -cost;    Edges[EdgeCount].next = Head[v];    Head[v] = EdgeCount++;}bool Spfa(int s, int t){    memset(Dist, 0x3f3f3f3f, sizeof(Dist));    memset(Pre, -1, sizeof(Pre));    Dist[s] = 0;    queue<int>Q;    Q.push(s);    while (!Q.empty()){        int u = Q.front();        Q.pop();        for (int e = Head[u]; e != -1; e = Edges[e].next){            int v = Edges[e].to;            if (Edges[e].vol > 0 && Dist[v] > Dist[u] + Edges[e].cost){                Dist[v] = Dist[u] + Edges[e].cost;                Pre[v] = u;                Path[v] = e;                Q.push(v);            }        }    }    return Pre[t] != -1;}int MCMF(int s, int t){    int cost = 0;    int max_flow = 0;    int u, v, e;    while (Spfa(s, t)){        int f = INF;        for (u = t; u != s; u = Pre[u]){            f = min(f, Edges[Path[u]].vol);        }        for (u = t; u != s; u = Pre[u]){            e = Path[u];            Edges[e].vol -= f;            Edges[e^1].vol += f;        }        max_flow += f;        cost += f*Dist[t];    }    if(max_flow == 2) return cost;    return -1;}void init(){    memset(Head,-1,sizeof(Head));    EdgeCount = 0;}int main(){    int n,m,Case = 1;    while(~scanf("%d%d",&n,&m)&&n+m){        init();        int u,v,cost;        while(m--){            scanf("%d%d%d",&u,&v,&cost);            addedge(u+1,v+1,1,cost);        }        addedge(0,1,2,0);        addedge(n,n+1,2,0);        printf("Instance #%d: ",Case++);        int cos = MCMF(0,n+1);        if(cos == -1) puts("Not possible");        else printf("%d\n",cos);    }    return 0;}