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;
}