Problem 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 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
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
题解
有上下界的网络流。这题是无源无汇的可行流的判断。
http://blog.sina.com.cn/s/blog_76f6777d0101bara.html这个网站上介绍的比较清楚。
大约思路是：假设每条路都有着“恰好等于下界”的流量，这时候会发现，有些路流进和流出的流量不相等，即不满足“流量平衡”。这时，对于那些“流进”>“流出”的边，从虚拟源点向“这条边的起点”连一条容量为“多出来的流入量”的边，对于 “流进”<“流出”的边的终点，类似地与虚拟汇点相连，然后将原本图上的边的容量变成“上界”-“下界”。
这样建图的目的是：在满足下界的情况下，在残量网络中“调整”所谓的“多出来的流”。让流量平衡。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<cmath>
#include<algorithm>
#define inf 1<<30
using namespace std;
int n,m,zz,head[210],in[210],totf,T;
struct bian{int to,nx,v;} e[100005];
int ans,h[210],q[210],bg[100005];
void insert(int x,int y,int z)
{
zz++; e[zz].to=y; e[zz].v=z; e[zz].nx=head[x]; head[x]=zz;
zz++; e[zz].to=x; e[zz].v=0; e[zz].nx=head[y]; head[y]=zz;
}
void init()
{
scanf("%d%d",&n,&m);
memset(in,0,sizeof(in));
memset(head,0,sizeof(head));
zz=1; totf=0; ans=0;
int i,x,y,l,r;
for(i=1;i<=m;i++)
{scanf("%d%d%d%d",&x,&y,&l,&r);
insert(x,y,r-l);
in[y]+=l; in[x]-=l;
bg[i]=r;
}
for(i=1;i<=n;i++)
{if(in[i]>0)
{insert(0,i,in[i]); totf+=in[i];}
else if(in[i]<0) insert(i,n+1,-in[i]);
}
}
bool bfs()
{
memset(h,-1,sizeof(h));
int i,x,p,t=0,w=1;
h[0]=0; q[0]=0;
while(t!=w)
{x=q[t]; t=(t+1)%205;
for(i=head[x];i;i=e[i].nx)
{p=e[i].to;
if(e[i].v&&h[p]<0)
{h[p]=h[x]+1;
q[w]=p; w=(w+1)%205;
}
}
}
if(h[n+1]==-1) return false;
else return true;
}
int dfs(int x,int f)
{
if(x==n+1) return f;
int i,rest,usd=0,p;
for(i=head[x];i;i=e[i].nx)
{p=e[i].to;
if(e[i].v&&h[p]==h[x]+1)
{rest=f-usd;
rest=dfs(p,min(e[i].v,rest));
e[i].v-=rest; e[i^1].v+=rest;
usd+=rest;
if(usd==f) return f;
}
}
if(usd==0) h[x]=-1;
return usd;
}
void dinic()
{
while(bfs()) ans+=dfs(0,inf);
if(ans!=totf) puts("NO");
else
{puts("YES");
int i;
for(i=1;i<=m;i++) printf("%d\n",bg[i]-e[i*2].v);
}
}
int main()
{
scanf("%d",&T);
while(T--)
{init(); dinic();}
return 0;
}