携程编程大赛1004

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define M 40009
#define N 2000
using namespace std;
struct edge{
    int y,next,f,d;
    bool vis;
}e[M];
int cnt,p[N],gap[N],h[N],n,m,source,sink,X,Y;
bool vis[N];
int dis[N];
void add(int a,int b,int d,int f){
    e[cnt].y=b;
    e[cnt].f=f;
    e[cnt].d=d;
    e[cnt].next=p[a];
    e[cnt].vis=0;
    p[a]=cnt++;
    e[cnt].y=a;
    e[cnt].d=d;
    e[cnt].f=f;
    e[cnt].next=p[b];
    e[cnt].vis=0;
    p[b]=cnt++;
}
void init(){
    memset(p,-1,sizeof(p));
    memset(h,0,sizeof(h));
    memset(gap,0,sizeof(gap));
    memset(vis,0,sizeof(vis));
    memset(dis,0xf,sizeof(dis));
    cnt=0;
    source=0;
    sink=n+1;
    gap[0]=n+2;//总点数
}
int dfs(int x,int f){
    if(x==sink)
        return f;
    int i,mh=n-1,lf=f,d,y;
    for(i=p[x];i!=-1;i=e[i].next){
        y=e[i].y;
        if(e[i].vis==0)
            continue;
        if(e[i].f>0){
            if(h[x]==h[y]+1){
                d=min(e[i].f,lf);
                d=dfs(y,d);
                e[i].f-=d;
                e[i^1].f+=d;
                lf-=d;
                if(h[source]>=n)
                    return f-lf;
                if(lf==0)
                    break;
            }
            mh=min(mh,h[y]);
        }
    }
    if(f==lf){
        gap[h[x]]--;
        if(gap[h[x]]==0)
            h[source]=n;
        h[x]=mh+1;
        gap[h[x]]++;
    }
    return f-lf;
}
int spfa(){
    queue<int>q;
    q.push(X);
    vis[X]=1;
    dis[X]=0;
    while(!q.empty()){
        int f=q.front();
        vis[f]=0;
        q.pop();
        for(int i=p[f];i!=-1;i=e[i].next){
            int y=e[i].y;
            if(dis[y]>dis[f]+e[i].d){
                dis[y]=dis[f]+e[i].d;
                if(!vis[y])
                    q.push(y),vis[y]=1;
            }
        }
    }
    return dis[Y];
}
void check(){
    queue<int>q;
    q.push(X);
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++){
        if(dis[i]>dis[Y])
            continue;
        for(int j=p[i];j!=-1;j=e[j].next){
            if(dis[e[j].y]>dis[Y]||dis[i]+e[j].d>dis[e[j].y])//判断边是否是最短路上的边
                continue;
            e[j].vis=1;
//            printf("x=%d y=%d f=%d\n",i,e[j].y,e[j].f);
        }
    }
}
int main(){
    int i,j,k,x,y,f,d;
    while(scanf("%d%d",&n,&m)&&n&&m){
        init();
        scanf("%d%d",&X,&Y);
        add(source,X,0,1e8);
        e[cnt-1].vis=1;
        e[cnt-2].vis=1;
        add(Y,sink,0,1e8);
        e[cnt-1].vis=1;
        e[cnt-2].vis=1;
        while(m--){
            scanf("%d%d%d%d",&x,&y,&d,&f);
            add(x,y,d,f);
        }
        spfa();
        check();
        int s=0;
        n+=2;
        while(h[source]<n)
            s+=dfs(source,1e8);
        printf("%d\n",s);
    }
    return 0;
}

先求出最短路，然后遍历每条边，如果边长大于两边到起点的距离只差，则该边不可走

将删除的代价作为流量，跑一边最大流（最小割）就是结果