算法复习——有源汇上下界可行流（bzoj2396）

Posted 2020-10-05 AseanA

题目：

Description

We are supposed to make a budget proposal for this multi-site competition. The budget proposal is a matrix where the rows represent different kinds of expenses and the columns represent different sites. We had a meeting about this, some time ago where we discussed the sums over different kinds of expenses and sums over different sites. There was also some talk about special constraints: someone mentioned that Computer Center would need at least 2000K Rials for food and someone from Sharif Authorities argued they wouldn\'t use more than 30000K Rials for T-shirts. Anyway, we are sure there was more; we will go and try to find some notes from that meeting. 

And, by the way, no one really reads budget proposals anyway, so we\'ll just have to make sure that it sums up properly and meets all constraints.

Input

The first line of the input contains an integer N, giving the number of test cases. The next line is empty, then, test cases follow: The first line of each test case contains two integers, m and n, giving the number of rows and columns (m <= 200, n <= 20). The second line contains m integers, giving the row sums of the matrix. The third line contains n integers, giving the column sums of the matrix. The fourth line contains an integer c (c < 1000) giving the number of constraints. The next c lines contain the constraints. There is an empty line after each test case. 

Each constraint consists of two integers r and q, specifying some entry (or entries) in the matrix (the upper left corner is 1 1 and 0 is interpreted as "ALL", i.e. 4 0 means all entries on the fourth row and 0 0 means the entire matrix), one element from the set {<, =, >} and one integer v, with the obvious interpretation. For instance, the constraint 1 2 > 5 means that the cell in the 1st row and 2nd column must have an entry strictly greater than 5, and the constraint 4 0 = 3 means that all elements in the fourth row should be equal to 3.

Output

For each case output a matrix of non-negative integers meeting the above constraints or the string "IMPOSSIBLE" if no legal solution exists. Put one empty line between matrices.

Sample Input

2

2 3 
8 10 
5 6 7 
4 
0 2 > 2 
2 1 = 3 
2 3 > 2 
2 3 < 5 

2 2 
4 5 
6 7 
1 
1 1 > 10

Sample Output

2 3 3 
3 3 4 

IMPOSSIBLE 

Source

Tehran 2003 Preliminary

题解：

算是求有源汇上下界最大流的一道模板题···只不过建图有点麻烦····

先说有源汇上下界最大流最小流的基本求法（引用自农民伯伯-Coding）：

有源汇网络的可行流

 对于流量有上下界的有源汇网络，原图中存在源点S和汇点T，为了求可行流，先将其转换为无源汇网络。 

从T-->S引入一条边，其流量上下界为[0, INF]，此时原图变为了无源汇网络，然后按照无源汇网络求解可行流的方法来求解。 

有源汇网络的最大流

 要求最大流，先求可行流，通过“有源汇网络的可行流”的求解方法来判断有源汇网络存在可行流。 
若存在可行流，记从S流出的流量sum1，然后将T-->S的边取消，再次从S到T求解网络的最大流，记从S流出的流量sum2. 那么该有源汇网络的最大流为 sum1 + sum2. 
其中，sum1是在网络满足流量下界的条件下，从源点S流出的流量；求出sum1之后，网络中可能还有余量可以继续增广，那么再次求解从S到T的最大流，得到sum2，sum1 + sum2即为最终的最大流。

有源汇网络的最小流

    求解有源汇网络最小流分为以下几步： 
（1）对SS到TT求一次最大流，即为f1.（在有源汇的情况下，先把整个网络趋向必须边尽量满足的情况）； 
（2）添加一条边T-->S，流量上限为INF，这条边即为P.(构造无源网络） 
（3）对SS到TT再次求最大流，即为f2。（判断此时的网络中存在可行流，同时求出SS-->TT最大流） 
    如果所有必须边都满流，证明存在可行解，原图的最小流为“流经边P的流量”（原图已经构造成无源汇网络，对于S同样满足 入流和==出流和，只有新添加的边流向S，而S的出流就是原图的最小流）。 
注： 最小流求法的正确性不知如何证明，待继续学习。

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=1000;
const int M=1000005;
const int inf=0x3f3f3f3f;
int tot=1,first[N],next[M],go[M],rest[M],lev[N],cur[N];
int tr[N],tc[N],n,m,low[N][N],up[N][N],T,src,des,ss,tt,sum1,sum2,k,ans,id[N][N],p;
bool jud[N][N];
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 comb1(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 int R()
{
  char c;
  int f=0;
  for(c=getchar();c<\'0\'||c>\'9\';c=getchar());
  for(;c<=\'9\'&&c>=\'0\';c=getchar())
    f=(f<<3)+(f<<1)+c-\'0\';
  return f;
}
inline void pre()
{
  memset(first,0,sizeof(first));
  memset(low,0,sizeof(low));
  memset(up,inf,sizeof(up));
  memset(tr,0,sizeof(tr));
  memset(tc,0,sizeof(tc));
  tot=1,src=0,des=n+m+1,ss=n+m+2,tt=n+m+3,sum1=0,sum2=0,ans=0;
}
inline bool bfs()
{
  for(int i=src;i<=tt;i++)  cur[i]=first[i],lev[i]=-1;
  static int que[N],tail,u,v;
  que[tail=1]=ss;
  lev[ss]=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==tt)  return true;
      }
    }
  }
  return false;
}
inline int dinic(int u,int flow)
{
  if(u==tt)
    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(ss,inf);
}
int main()
{
  //freopen("a.in","r",stdin);
  T=R();
  int a,b,c;
  char s[5];
  while(T--)
  {
    scanf("\\n");
    n=R(),m=R();
    pre();
    for(int i=1;i<=n;i++)
    {
      a=R();
      tc[src]+=a,tr[i]+=a;  
      sum1+=a;
    }
    for(int i=1;i<=m;i++)
    {
      a=R();
      tr[des]+=a,tc[n+i]+=a;
      sum2+=a;
    }
    if(sum1!=sum2)  
    {  
      cout<<"IMPOSSIBLE"<<endl;
      cout<<endl;
      continue;
    }
    k=R();    
    while(k--)
    {
      scanf("%d%d%s%d",&a,&b,s,&c);
      if(a==0&&b==0)
      {
        for(int i=1;i<=n;i++)
          for(int j=1;j<=m;j++)
          {
            if(s[0]==\'=\')
            {  
              low[i][j]=max(low[i][j],c);
              up[i][j]=min(up[i][j],c);
            }
            if(s[0]==\'<\')  
              up[i][j]=min(up[i][j],c-1);
            if(s[0]==\'>\')
              low[i][j]=max(low[i][j],c+1);
          }
      }
      else if(a==0)
        for(int i=1;i<=n;i++)
        {
          if(s[0]==\'=\')
          {
            low[i][b]=max(low[i][b],c);
            up[i][b]=min(up[i][b],c);
          }
          if(s[0]==\'<\')
            up[i][b]=min(up[i][b],c-1);
          if(s[0]==\'>\')
            low[i][b]=max(low[i][b],c+1);
        }     
      else if(b==0)
        for(int i=1;i<=m;i++)
        {
          if(s[0]==\'=\')
          {
            low[a][i]=max(low[a][i],c);
            up[a][i]=min(up[a][i],c);
          }
          if(s[0]==\'<\')
            up[a][i]=min(up[a][i],c-1);
          if(s[0]==\'>\')
            low[a][i]=max(low[a][i],c+1);
        }
      else
      {
        if(s[0]==\'=\')
          {
            low[a][b]=max(low[a][b],c);
            up[a][b]=min(up[a][b],c);
          }
        if(s[0]==\'<\')
          up[a][b]=min(up[a][b],c-1);
        if(s[0]==\'>\')
          low[a][b]=max(low[a][b],c+1); 
      }
    }
    bool flag=false;
    for(int i=1;i<=n;i++)
    {  
      if(flag==true)  break;
      for(int j=1;j<=m;j++)
      {
        if(low[i][j]>up[i][j]) 
        { 
          flag=true;
          break;
        }
        else
        {
          tr[n+j]+=low[i][j];
          tc[i]+=low[i][j];
          id[i][j]=++tot;
          comb1(i,n+j,up[i][j]-low[i][j]);
        }
      }
    }
    if(flag==true)
    {
      cout<<"IMPOSSIBLE"<<endl;
      cout<<endl;
      continue;
    }
    int cnt=0;
    p=++tot;
    comb1(des,src,inf);
    for(int i=src;i<=des;i++)
    {
      if(tr[i]>tc[i])
      {  
        comb(ss,i,tr[i]-tc[i]);
        cnt+=tr[i]-tc[i];
      }
      if(tc[i]>tr[i])
        comb(i,tt,tc[i]-tr[i]);
    }
    maxflow();
    if(ans!=cnt)
    {
      cout<<"IMPOSSIBLE"<<endl;
      cout<<endl;
      continue;
    }
    rest[p]=rest[p^1]=0;
    maxflow();  
    for(int i=1;i<=n;i++)
    {  
      for(int j=1;j<=m;j++)
        cout<<rest[id[i][j]^1]+low[i][j]<<" ";
      cout<<endl;
    }
    cout<<endl;
  }
  return 0;
}