模板——最小费用流

#include<bits/stdc++.h>
using namespace std;
const int P=1000000007;
struct nob{
    int to,jump,val,cost;
}a[500000];
int n,m,s,t,tot=1,jump[500000]={0};
int len[10005],line[1000005];
bool judge[10005];
inline int read(){
    int rem=0,f=1;
    char cha=getchar();
    while(cha>'9'||cha<'0'){
        if(cha=='-'){
            f=-1;
        }
        cha=getchar();
    } 
    while(cha<='9'&&cha>='0'){
        rem=rem*10+cha-'0';
        cha=getchar();
    }
    return rem*f;
}
void add(int x,int y,int z,int d){
    tot++;
    a[tot].to=y;
    a[tot].jump=jump[x];
    jump[x]=tot;
    a[tot].cost=d;
    a[tot].val=z;
}
bool SPFA(){
    int head=0,tail=1,pos;
    memset(judge,0,sizeof(judge));
    for (int i=1; i<=n; i++) len[i]=P;
    len[s]=0;
    judge[s]=1;
    line[1]=s;
    while (head!=tail){
        head++;
        if (head>1000000) head=1;
        pos=line[head];
        for (int i=jump[pos]; i; i=a[i].jump){
            if (a[i].val && len[pos]+a[i].cost<len[a[i].to]){
                len[a[i].to]=len[pos]+a[i].cost;
                if (!judge[a[i].to]){
                    judge[a[i].to]=1;
                    tail++;
                    if (tail>1000000) tail=1;
                    line[tail]=a[i].to;
                }
            }
        }
        judge[pos]=0;
    }
    if (len[t]==P) return 0;
    else return 1;
}
int vis[8000+10];
int cnt=0;
int dfs(int pos,int l){
    if (pos==t) return l;
    int used=0,w=0;
    vis[pos]=cnt;
    for (int i=jump[pos]; i; i=a[i].jump){
        if (vis[a[i].to]!=cnt && len[a[i].to]==len[pos]+a[i].cost && a[i].val){
            w=l-used;
            int le=dfs(a[i].to,min(w,a[i].val));
            a[i].val-=le;
            a[i^1].val+=le;
            used+=le;
            if (used==l) return l;
        }
    }
    return used;
}
int main(){
    int maxn=0,ans=0;
    n=read();m=read();s=read();t=read();
    for (int i=1,x,y,z,d; i<=m; i++){
        x=read();y=read();z=read();d=read();
        add(x,y,z,d);
        add(y,x,0,-d);
    }
    int p;
    while (SPFA()){
    	++cnt;
        p=dfs(s,P);
        maxn+=p;
        ans+=p*len[t];
    }
    printf("%d %d",maxn,ans);
    return 0;
}

  

posted @ 2018-03-11 16:36 |斗蜂| 阅读(...) 评论(...) 编辑 收藏