uva 1658 Admiral 最小費最大流

題意就是讓你求兩次1到n的最短路。這題應該可以用最短路來求解吧，只需要將第一次用到的邊刪去即可。我這裡是按照算法競賽入門經典裡面提到拆點+最小費最大流。

#include
using namespace std;
const int N=1024*4;
const int inf=1<<24;

struct Edge
{
    int from,to,cap,flow,cost;
};
vectoredges;
vectorG[N];
int n,m;
int inq[N],p[N],d[N],a[N];

void AddEdge(int from, int to,int cap, int cost)
{
    Edge tp;
    tp.from=from,tp.to=to,tp.cap=cap,tp.flow=0,tp.cost=cost;
    edges.push_back(tp);

    tp.from=to,tp.to=from,tp.cap=0,tp.flow=0,tp.cost=-cost;
    edges.push_back(tp);

    int g=edges.size();
    G[from].push_back(g-2);
    G[to].push_back(g-1);
}

int BellmanFord(int s,int t,int &flow, int &cost)
{
    int i,j,u;
    for(i=0; iQ;
    Q.push(s);
    while(!Q.empty())
    {

        u=Q.front();
        Q.pop();
        inq[u]=0;
        for(i=0; ie.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 0;
    flow+=a[t];
    cost+=d[t]*a[t];
    u=t;
    while(u!=s)
    {
        edges[p[u]].flow+=a[t];
        edges[p[u]^1].flow-=a[t];
        u=edges[p[u]].from;
    }
    return 1;
}

int Mincost(int s,int t)
{
    int flow=0,cost=0;
    BellmanFord(s,t,flow,cost);
    return cost;
}

int main()
{
    int i,u,v,c;
    while(~scanf(%d%d,&n,&m))
    {
        for(i=0; i