zoj 2314 Reactor Cooling （无源汇上下界可行流）

Reactor Coolinghttp://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1314

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

m根管道，构成一个循环体，每时每刻 管道要满足 流进的=流出的

每根管道有流量下限和上限

求是否能够成这样一个循环体，如果能，输出每条管道的流量

（不知道是不是最大，反正他们都是最大。。。）

无源汇上下界可行流的模板题

推荐一篇写的相当好的博客http://www.cnblogs.com/liu-runda/p/6262832.html

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define N 250
#define M 100000
using namespace std;
int a[N],low[M],sum,tot;
int front[N],next[M],to[M],cap[M];
int src,dec;
int cur[N],lev[N];
queue<int>q;
void add(int u,int v,int w)
{
    to[++tot]=v; next[tot]=front[u]; front[u]=tot; cap[tot]=w;
    to[++tot]=u; next[tot]=front[v]; front[v]=tot; cap[tot]=0;
}
bool bfs()
{
    for(int i=src;i<=dec;i++) lev[i]=-1,cur[i]=front[i];
    while(!q.empty()) q.pop();
    lev[src]=0;
    q.push(src);
    int now;
    while(!q.empty())
    {
        now=q.front(); q.pop();
        for(int i=front[now];i;i=next[i])
         if(cap[i]>0&&lev[to[i]]==-1)
         {
              lev[to[i]]=lev[now]+1;
              if(to[i]==dec) return true;
              q.push(to[i]);
         }
    }
    return false;
}
int dfs(int now,int flow)
{
    if(now==dec) return flow;
    int rest=0,delta;
    for(int & i=cur[now];i;i=next[i])
     if(cap[i]>0&&lev[to[i]]>lev[now])
     {
          delta=dfs(to[i],min(flow-rest,cap[i]));
          if(delta)
          {
                cap[i]-=delta; cap[i^1]+=delta;
              rest+=delta; if(rest==flow) break;
          }
     }
     if(rest!=flow) lev[now]=-1;
     return rest;
}
int dinic()
{
    int tmp=0;
    while(bfs())
    tmp+=dfs(src,2e9);
    return tmp;
}
int main()
{
    int T ,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        src=0,dec=n+1;
        tot=1; sum=0;
        memset(front,0,sizeof(front));
        memset(a,0,sizeof(a));
        int f,t,u,d;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d%d",&f,&t,&u,&d);
            low[i]=u;
            a[f]-=u; a[t]+=u;
            add(f,t,d-u);
        }
        for(int i=1;i<=n;i++)
        if(a[i]<0) add(i,dec,-a[i]);
        else if(a[i]>0) add(src,i,a[i]),sum+=a[i];
        if(dinic()==sum)
        {
            puts("YES");
            for(int i=1;i<=m;i++) printf("%d\n",low[i]+cap[i<<1|1]);
        }
        else puts("NO");
    }
}