LOJ155 「模板」无源汇有上下界可行流 上下界网络流

问题描述

$ n $ 个点，$ m $ 条边，每条边 $ e $ 有一个流量下界 $ \text{lower}(e) $ 和流量上界 $ \text{upper}(e) $，求一种可行方案使得在所有点满足流量平衡条件的前提下，所有边满足流量限制。

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https://www.luogu.com.cn/blog/duyi/qian-tan-you-shang-xia-jie-di-wang-lao-liu

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\(\mathrm{Code}\)

#include<bits/stdc++.h>
using namespace std;

const int maxn=220;
const int maxm=500007;
const int INF=0x3f3f3f3f;

int n,m,S,T;
int Head[maxn],to[maxm],Next[maxm],w[maxm],tot=1;
int Getin[maxn],Sendout[maxn];

int Low[maxm];

void addedge(int x,int y,int z){
    to[++tot]=y,Next[tot]=Head[x],Head[x]=tot,w[tot]=z;
}
void add(int x,int y,int z){
    addedge(x,y,z);addedge(y,x,0);
}

int d[maxn];

bool bfs(void){
    memset(d,0,sizeof(d));
    queue<int>q;q.push(S);d[S]=1;
    while(!q.empty()){
        int x=q.front();q.pop();
        for(int i=Head[x];i;i=Next[i]){
            int y=to[i];
            if(d[y]||!w[i]) continue;
            d[y]=d[x]+1;q.push(y);
            if(y==T) return true;
        }
    }
    return false;
}

int dfs(int x,int flow){
    if(x==T) return flow;
    int rest=flow;
    for(int i=Head[x];i;i=Next[i]){
        int y=to[i];
        if(d[y]!=d[x]+1||!w[i]) continue;
        int k=dfs(y,min(rest,w[i]));
        if(!k) d[y]=0;
        else w[i]-=k,w[i^1]+=k,rest-=k;
    }
    return flow-rest;
}

int MaxFlow;

void Dinic(void){
    int t;
    while(bfs()){
        while(t=dfs(S,INF)) MaxFlow+=t;
    }
}

void Init(void){
    scanf("%d%d",&n,&m);
    for(int i=1,x,y,low,upp;i<=m;i++){
        scanf("%d%d%d%d",&x,&y,&low,&upp);
        Getin[y]+=low,Sendout[x]+=low;
        add(x,y,upp-low);Low[i]=low;
    }
}

bool check(void){
    for(int i=Head[S];i;i=Next[i]){
        if(w[i]) return false;
    }
    return true;
}

void Work(void){
    S=n+1,T=S+1;
    for(int i=1;i<=n;i++){
        int diff=Getin[i]-Sendout[i];
        if(!diff) continue;
        if(diff>0) add(S,i,diff);
        else add(i,T,-diff);
    }
    Dinic();
    if(!check()){
        puts("NO");return;
    }
    puts("YES");
    for(int i=2;i<=2*m+1;i+=2){
        printf("%d\n",w[i^1]+Low[(i+1)/2]);
    }
}

int main(){
    Init();
    Work();
    return 0;
}