ZOJ [P2314] 无源汇点有上下界模版

对于有上下界的网络流来说,我们可以分离出必要弧,然后将必要弧切开,两端分别连接源点和汇点,原图有可行解充要于源点或汇点满流.
这样求下来,只能求出可行流

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <queue>
using namespace std;
int n,m,s,t,head[250],cur[250],dep[250],nume;
int init(){
	int rv=0,fh=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-') fh=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		rv=(rv<<1)+(rv<<3)+c-'0';
		c=getchar();
	}
	return fh*rv;
}
struct edge{
	int to,nxt,cap,flow;
}e[500005];
void adde(int from,int to,int cap){
	e[++nume].to=to;
	e[nume].cap=cap;
	e[nume].nxt=head[from];
	head[from]=nume;
	e[nume].flow=0;
}
bool bfs(){
	queue<int> q;
	while(!q.empty()) q.pop();
	memset(dep,0,sizeof(dep));
	q.push(s);
	dep[s]=1;
	while(!q.empty()){
		int u=q.front();q.pop();
		for(int i=head[u];i;i=e[i].nxt){
			int v=e[i].to;
			if(!dep[v]&&(e[i].flow<e[i].cap)){
				dep[v]=dep[u]+1;
				q.push(v);
			}
		}
	}
	if(dep[t])	return 1;
	else return 0;
}
int dfs(int u,int flow){
	if(u==t) return flow;
	int tot=0;
	for(int &i=cur[u];i&&tot<flow;i=e[i].nxt){
		int v=e[i].to;
		//printf("%d %d\n",v,e[i].cap-e[i].flow);
		if((dep[v]==dep[u]+1)&&(e[i].flow<e[i].cap)){
			if(int t=dfs(v,min(flow-tot,e[i].cap-e[i].flow))){	
				e[i].flow+=t;
				//cout<<t<<endl;
				e[((i-1)^1)+1].flow-=t;
				tot+=t;
			}
		}
	}
	return tot;
}
int dinic(){
	int ans=0;
	while(bfs()){
		for(int i=1;i<=n+2;i++) cur[i]=head[i];
		ans+=dfs(s,0x3f3f3f3f);	
	}
	return ans;
}
int main(){
	int T=init();
	while(T--){
		int tot=0,ans=0;
		n=init();m=init();
		s=n+2;t=n+1;
		memset(head,0,sizeof(head));
		nume=0;
		for(int i=1;i<=m;i++){
			int u=init(),v=init(),b=init(),d=init();
			adde(u,v,d-b);
			adde(v,u,0);
			adde(s,v,b);//千万不要建反
			adde(v,s,0);
			adde(u,t,b);
			adde(t,u,0);
			tot+=b;
		}
		ans=dinic();
	//	for(int i=head[s];i;i=e[i].nxt) cout<<e[i].to<<endl;
		if(ans<tot) printf("NO\n");
		else{
			printf("YES\n");
			for(int i=1;i<=nume;i+=6){
				printf("%d\n",e[i].flow+e[i+2].cap);
			}
		}
		printf("\n");
	}
	return 0;
}