UVA1658：Admiral

题意：给定一个有向带权图，求两条不相交（无公共点）的路径且路径权值之和最小，路径由1到v

题解：这题的关键就在于每个点只能走一遍，于是我们想到以边换点的思想，用边来代替点，怎么代替呢？

把i拆成i和i'，这样经过i就转化为经过i到i'的路径了，从而用最小费用流即可

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#define MAXN 5005
#define ll long long
#define INF 0x7f7f7f7f
using namespace std;
struct Edge{
    int from,to,cap,flow,cost;
    Edge(int u=0,int v=0,int c=0,int f=0,int w=0){
        from=u,to=v,cap=c,flow=f,cost=w;
    }
};
struct MCMF{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[MAXN];
    int d[MAXN];
    int p[MAXN];
    int b[MAXN];
    int a[MAXN];
    void init(int n,int s,int t){
        this->n=n;
        this->s=s,this->t=t;
        edges.clear();
        for(int i=0;i<=n;i++){
            G[i].clear();
        }
    }
    void AddEdge(int x,int y,int cap,int cost){
        edges.push_back(Edge(x,y,cap,0,cost));
        edges.push_back(Edge(y,x,0,0,-cost));
        m=edges.size();
        G[x].push_back(m-2);
        G[y].push_back(m-1);
    }
    int SPFA(int &flow,ll &cost){
        memset(d,0x7f,sizeof(d));
        memset(b,0,sizeof(b));
        queue<int> q;
        p[s]=0;
        a[s]=INF;
        d[s]=0;
        q.push(s);
        b[s]=1;
        while(!q.empty()){
            int x=q.front();q.pop();
            b[x]=0;
            for(int i=0;i<G[x].size();i++){
                Edge& e=edges[G[x][i]];
                if(e.cap>e.flow&&d[e.to]>d[x]+e.cost){
                    p[e.to]=G[x][i];
                    a[e.to]=min(a[x],e.cap-e.flow);
                    d[e.to]=d[x]+e.cost;
                    if(!b[e.to]){
                        b[e.to]=1;
                        q.push(e.to);
                    }
                }
            }
        }
        if(d[t]==INF){
            return 0;
        }
        flow+=a[t];
        cost+=1LL*d[t]*a[t];
        for(int i=t;i!=s;i=edges[p[i]].from){
            edges[p[i]].flow+=a[t];
            edges[p[i]^1].flow-=a[t];
        }
        return 1;
    }
    ll MincostMaxflow(){
        int flow=0;ll cost=0;
        while(SPFA(flow,cost));
        return cost;
    }
}D;
int n,m;
int main()
{
    while(~scanf("%d%d",&n,&m)){
        D.init(n<<1,0,n);
        D.AddEdge(0,1,2,0);
        for(int i=1;i<=n;i++){
            D.AddEdge(i,i+n,1,0);
        }
        for(int i=1;i<=m;i++){
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            if(x==1||x==n)
                D.AddEdge(x,y,1,1LL*w);
            else
                D.AddEdge(x+n,y,1,1LL*w);
        }
        printf("%lld
",D.MincostMaxflow());
    }
    return 0;
}