无源汇有上下界可行流存在定理

H - Reactor Cooling

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status

Description

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 f_ij, (put f_ij = 0 if there is no pipe from node i to node j), for each i the following condition must hold:

f _i,1+f _i,2+...+f _i,N = f _1,i+f _2,i+...+f _N,i

Each pipe has some finite capacity, therefore for each i and j connected by the pipe must be f_ij <= c_ij 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 l_ij, thus it must be f_ij >= l_ij.

Given cij and l_ij for all pipes, find the amount f_ij, 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 <= l_ij<= c_ij <= 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

2

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

NO

YES

1

2

3

2

1

1

这是一个无源汇有上下界处理是否存在可行流的网络最大流改进模型；

一个可行流的概念是    1）Ef(u,i) ==Ef(i,v)&&f(u,v)<=C(u,v)，而有上下界的可行流概念是   2）Ef(u,i)
==Ef(i,v)&&B(u,v)<=f(u,v)<=C(u,v),因此将上下界的图转化为无下界的图。但在转化的时候，我们需要创造条件1，因此需要利用下界的条件进行推导：

http://wenku.baidu.com/view/0f3b691c59eef8c75fbfb35c.html

看懂上面的，这个题其实是个模板：

#include<stdio.h>
#include<string.h>
#include<queue>
#define INF 0x7fffffff
#define size 250
using namespace std;
struct node{
	int a,b,aa;
}vv[size];
queue<int>Q;
int map[size][size],pre[size];
int bfs(int s,int e){
	int flow[250];
	for(int i=s;i<=e;i++)pre[i] = -1;
	pre[s] = 0;
	while(!Q.empty())Q.pop();
	Q.push(s);
	flow[s] = INF;
	while(!Q.empty()){
		int a= Q.front();
		Q.pop();
		if(a==e)break;
		for(int i=s;i<=e;i++){
			if(map[a][i]>0&&pre[i]==-1){
				Q.push(i);
				pre[i] = a;
				flow[i] = min(flow[a],map[a][i]);
			}
		}
	}
	return pre[e]==-1?-1:flow[e];
}
int main(){
	int t;
	int ds[250][250],ru[250];
	int flow;
	scanf("%d",&t);
	while(t--){
		getchar();
		memset(ru,0,sizeof(ru));
		memset(map,0,sizeof(map));
		memset(ds,0,sizeof(ds));
		int m,n,a,b,aa,bb;
		scanf("%d%d",&n,&m);
		for(int i=0;i<m;i++){
			scanf("%d%d%d%d",&a,&b,&aa,&bb);
			map[a][b] = bb-aa;
			ru[b]+=aa;
			ru[a]-=aa;
			vv[i].a = a;
			vv[i].b = b;
			vv[i].aa = aa;
		}
		for(int i=1;i<=n;i++)
		if(ru[i]>=0)map[0][i] = ru[i];
		else map[i][n+1] = -1*ru[i];
		while((flow = bfs(0,n+1))!=-1){
			int k = n+1;
			while(k!=0){
				int last;
				last = pre[k];
				map[last][k] -=flow;
				map[k][last] +=flow;
				ds[last][k] +=1;
				k = last;
			}
		}
		int i;
		//for(int i=0;i<=n+1;i++)printf("map1 = %d map2 = %d ds1 = %d ds2 = %d\n",map[0][i],map[i][n+1],ds[0][i],ds[i][n+1]);
		for(i=0;i<=n+1;i++)
		if(map[0][i]==0&&map[i][n+1]==0);
		else break;
		if(i==n+2){
			puts("YES");
			for(i=0;i<m;i++)
			printf("%d\n",ds[vv[i].a][vv[i].b]+vv[i].aa);
		}
		else puts("NO");
		if(t!=0)printf("\n");
	}
} 

无源汇有上下界可行流存在定理