Reactor Cooling_牛客博客

The terrorist group leaded by a well known international terrorist Ben Bladen is buliding a nuclear reactor to produce plutonium for the nuclear bomb they are planning to create. Being the wicked computer genius of this group, you are responsible for developing the cooling system for the reactor.

The cooling system of the reactor consists of the number of pipes that special cooling liquid flows by. Pipes are connected at special points, called nodes, each pipe has the starting node and the end point. The liquid must flow by the pipe from its start point to its end point and not in the opposite direction.

Let the nodes be numbered from 1 to N. The cooling system must be designed so that the liquid is circulating by the pipes and the amount of the liquid coming to each node (in the unit of time) is equal to the amount of liquid leaving the node. That is, if we designate the amount of liquid going by the pipe from i-th node to j-th as fij, (put fij = 0 if there is no pipe from node i to node j), for each i the following condition must hold:

fi,1+fi,2+…+fi,N = f1,i+f2,i+…+fN,i
Each pipe has some finite capacity, therefore for each i and j connected by the pipe must be fij <= cij where cij is the capacity of the pipe. To provide sufficient cooling, the amount of the liquid flowing by the pipe going from i-th to j-th nodes must be at least lij, thus it must be fij >= lij.

Given cij and lij for all pipes, find the amount fij, satisfying the conditions specified above.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

The first line of the input file contains the number N (1 <= N <= 200) - the number of nodes and and M - the number of pipes. The following M lines contain four integer number each - i, j, lij and cij each. There is at most one pipe connecting any two nodes and 0 <= lij <= cij <= 10^5 for all pipes. No pipe connects a node to itself. If there is a pipe from i-th node to j-th, there is no pipe from j-th node to i-th.

Output

On the first line of the output file print YES if there is the way to carry out reactor cooling and NO if there is none. In the first case M integers must follow, k-th number being the amount of liquid flowing by the k-th pipe. Pipes are numbered as they are given in the input file.

Sample Input

4 6
1 2 1 2
2 3 1 2
3 4 1 2
4 1 1 2
1 3 1 2
4 2 1 2

4 6
1 2 1 3
2 3 1 3
3 4 1 3
4 1 1 3
1 3 1 3
4 2 1 3

Sample Input

YES
1
2
3
2
1
1

一道无汇源上下界可行流。

具体建图：

原图每条边的上下界为l,r ，那么新图的此边的流量为 r-l ，并且记录每个点的流入下和界和流出下界和。
对于每个点的流出下界和，若大于0则让S连向此点流量为和，若小于0则连向T，流量为和。
跑S到T的最大流，若为S流出之和则有解，否则无解。
输出每条边的实际流量：下界+反向边的流量。

AC代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=210,M=1e6+10,inf=0x3f3f3f3f;
int T,n,m,d[N],h[N],ss,tt=201,l[M],sum;
int head[N],nex[M],to[M],w[M],tot,cur[N];
inline void ade(int a,int b,int c){
	to[++tot]=b; nex[tot]=head[a]; w[tot]=c; head[a]=tot;
}
inline void add(int a,int b,int c){
	ade(a,b,c); ade(b,a,0);
}
inline void init(){
	tot=1; memset(head,0,sizeof head); memset(d,0,sizeof d); sum=0;
}
inline int bfs(){
	queue<int> q; q.push(ss); memset(h,0,sizeof h); h[ss]=1;
	while(q.size()){
		int u=q.front(); q.pop();
		for(int i=head[u];i;i=nex[i]){
			if(!h[to[i]]&&w[i]){
				h[to[i]]=h[u]+1;	q.push(to[i]); 
			}
		}
	}
	return h[tt];
}
int dfs(int x,int f){
	if(x==tt)	return f; int fl=0;
	for(int i=head[x];i&&f;i=nex[i]){
		if(w[i]&&h[to[i]]==h[x]+1){
			int mi=dfs(to[i],min(f,w[i]));
			w[i]-=mi; w[i^1]+=mi; fl+=mi; f-=mi;
		}
	}
	if(!fl) h[x]=-1;
	return fl;
}
int dinic(){
	int res=0;
	while(bfs())	res+=dfs(ss,inf);
	return res;
}
signed main(){
	scanf("%d",&T);
	while(T--){	
		scanf("%d %d",&n,&m);	init();
		for(int i=1;i<=m;i++){
			int a,b,r;	scanf("%d %d %d %d",&a,&b,&l[i],&r);
			add(a,b,r-l[i]);	d[a]-=l[i];	d[b]+=l[i]; 
		}
		for(int i=1;i<=n;i++){
			if(d[i]>0)	add(ss,i,d[i]),sum+=d[i];
			else if(d[i]<0)	add(i,tt,-d[i]);
		}
		if(dinic()==sum){
			puts("YES");
			for(int i=1;i<=m;i++){
				printf("%d\n",l[i]+w[(i<<1)^1]);
			}
		}else	puts("NO");
	}
	return 0;
}