UVa1658 Admiral

题目大意：给出一个v个点e条边的有向加权图，求1到v的两条不相交的（除了起点和终点没有公共点）的路径使边权之和最小。
我有话说：
用拆点法，把一个点分成两个点，中间连一条边，边的容量为1，费用为0，然后求1到v的流量为2的最小费用流即可。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <cassert>#include <queue>using namespace std;const int maxn=2000+10;const int INF=1000000000;struct Edge{  int from,to,cap,flow,cost;  Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}};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;//无法增广    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);//每个点拆成两个点分编号为i-1和i+n-2（其实是i和i+n-1）      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;}