算法复习——无源汇可行流（zoj2314）

题目：

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

题解：

先说说无源汇可行流的解法：

计算每个顶点的r和c，其中r表示进入该点的边的下界值之和，c表示从该点出发的边的下界值之和

若r>c,则将该点与src（源点）连一条下界为0，上界为r-c的边

若r<c,则将该点与des（汇点）连一条下界为0，上界为c-r的边

然后原来的边怎么连就怎么连，但下界改为0,上界为这条边的原来的上界减去原来的下界

然后跑最大流，若从src出发的边都跑满则有界，否则无解

边的实际流量就是这条边原来的下界加上此时边的流量

该题为无源汇可行流的模板题

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cctype>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int N=400;
const int M=100005;
int T,n,m,des,src,tr[N],tc[N],id[M];
int tot=1,first[N],lev[N],go[M],next[M],rest[M],cur[M],cnt=0,ans=0;
struct node
{
  int from,go,minn,maxx;
}edge[M];
inline void comb(int a,int b,int c)
{
  next[tot]=first[a],first[a]=tot,go[tot]=b,rest[tot]=c;
  next[++tot]=first[b],first[b]=tot,go[tot]=a,rest[tot]=0;
}
inline void comb2(int a,int b,int c)
{
  next[++tot]=first[a],first[a]=tot,go[tot]=b,rest[tot]=c;
  next[++tot]=first[b],first[b]=tot,go[tot]=a,rest[tot]=0;
}
inline bool bfs()
{
  for(int i=src;i<=des;i++)  cur[i]=first[i],lev[i]=-1;
  static int que[N],tail,u,v;
  que[tail=1]=src;
  lev[src]=0;
  for(int head=1;head<=tail;head++)
  {
    u=que[head];
    for(int e=first[u];e;e=next[e])
    {
      if(lev[v=go[e]]==-1&&rest[e])
      {
        lev[v]=lev[u]+1;
        que[++tail]=v;
        if(v==des)  return true;
      }
    }
  }
  return false;
}
inline int dinic(int u,int flow)
{
  if(u==des)
    return flow;
  int res=0,delta,v;
  for(int &e=cur[u];e;e=next[e])
  {
    if(lev[v=go[e]]>lev[u]&&rest[e])
    {
      delta=dinic(v,min(flow-res,rest[e]));
      if(delta)
      {
        rest[e]-=delta;
        rest[e^1]+=delta;
        res+=delta;
        if(res==flow)  break;
      }
    }
  }
  if(flow!=res)  lev[u]=-1;
  return res;
}
inline void maxflow()
{
  while(bfs())
    ans+=dinic(src,1e+8);
}
int main()
{ 
  //freopen("a.in","r",stdin);
  scanf("%d",&T);
  while(T--)  
  {    
    scanf("
");
    memset(tr,0,sizeof(tr));
    memset(tc,0,sizeof(tc));
    memset(first,0,sizeof(first));
    scanf("%d%d",&n,&m);
    src=0,des=n+1,tot=1,cnt=0,ans=0;
    for(int i=1;i<=m;i++)
    {
      scanf("%d%d%d%d",&edge[i].from,&edge[i].go,&edge[i].minn,&edge[i].maxx);
      tr[edge[i].go]+=edge[i].minn;
      tc[edge[i].from]+=edge[i].minn;
      id[i]=++tot;                                                                 
      comb(edge[i].from,edge[i].go,edge[i].maxx-edge[i].minn);
    }
    for(int i=1;i<=n;i++)
    {
      if(tr[i]>tc[i])
      {  
        comb2(src,i,tr[i]-tc[i]);
        cnt+=(tr[i]-tc[i]);
      }
      if(tr[i]<tc[i])
        comb2(i,des,tc[i]-tr[i]);
    }
    maxflow();
    if(ans!=cnt)  cout<<"NO"<<endl;
    else
    {  
      cout<<"YES"<<endl;
      for(int i=1;i<=m;i++)
        cout<<(rest[id[i]^1]+edge[i].minn)<<endl; 
    }
  }
}