HDOJ 4687 Boke and Tsukkomi 一般图最大匹配带花树+暴力

一般图最大匹配带花树+暴力:

先算最大匹配 C1

在枚举每一条边,去掉和这条边两个端点有关的边.....再跑Edmonds得到匹配C2

如果C2+2==C1则这条边再某个最大匹配中

Boke and Tsukkomi

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 649    Accepted Submission(s): 202

Problem Description

A new season of Touhou M-1 Grand Prix is approaching. Girls in Gensokyo cannot wait for participating it. Before the registration, they have to decide which combination they are going to compete as. Every girl in Gensokyo is both a boke (funny girl) and a tsukkomi (straight girl). Every candidate combination is made up of two girls, a boke and a tsukkomi. A girl may belong to zero or more candidate combinations, but one can only register as a member of one formal combination. The host of Touhou M-1 Grand Prix hopes that as many formal combinations as possible can participate in this year. Under these constraints, some candidate combinations are actually redundant as it\'s impossible to register it as a formal one as long as the number of formal combinations has to be maximized. So they want to figure out these redundant combinations and stop considering about them.

 

Input

There are multiple test cases. Process to the End of File.
The first line of each test case contains two integers: 1 ≤ N ≤ 40 and 1 ≤ M ≤  123, where N is the number of girls in Gensokyo, and M is the number of candidate combinations. The following M lines are M candidate combinations, one by each line. Each combination is represented by two integers, the index of the boke girl 1 ≤ B _i ≤ N and the index of the tsukkomi girl 1 ≤ T _i ≤ N, where B _i != T _i.

 

Output

For each test case, output the number of redundant combinations in the first line. Then output the space-separated indexes of the redundant combinations in ascending order in the second line.

 

Sample Input

   
   
   
   

    
    
    
    4 4
1 3
2 3
2 4
3 1
6 6
1 2
3 2
3 4
5 2
5 4
5 6
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
3
2 4 5
   
   
   
   

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

const int maxn=50;

vector<int> ans;

bool G[maxn][maxn],TG[maxn][maxn];

int n,m;
int Match[maxn];
int Start,Finish,NewBase;
int Father[maxn],Base[maxn];
bool InQueue[maxn],InPath[maxn],InBlossom[maxn];
int Count;
queue<int> q;

int FindCommonAncestor(int u,int v)
{
  memset(InPath,false,sizeof(InPath));
  while(true)
    {
      u=Base[u];
      InPath[u]=true;
      if(u==Start) break;
      u=Father[Match[u]];
    }
  while(true)
    {
      v=Base[v];
      if(InPath[v]) break;
      v=Father[Match[v]];
    }
  return v;
}

void ResetTrace(int u)
{
  int v;
  while(Base[u]!=NewBase)
    {
      v=Match[u];
      InBlossom[Base[u]]=InBlossom[Base[v]]=true;
      u=Father[v];
      if(Base[u]!=NewBase) Father[u]=v;
    }
}

void BlossomContract(int u,int v)
{
  NewBase=FindCommonAncestor(u,v);
  memset(InBlossom,false,sizeof(InBlossom));
  ResetTrace(u); ResetTrace(v);
  if(Base[u]!=NewBase) Father[u]=v;
  if(Base[v]!=NewBase) Father[v]=u;
  for(int tu=1;tu<=n;tu++)
    {
      if(InBlossom[Base[tu]])
        {
          Base[tu]=NewBase;
          if(!InQueue[tu])
            {
              q.push(tu);
              InQueue[tu]=true;
            }
        }
    }
}

void FindAugmentingPath()
{
  memset(InQueue,false,sizeof(InQueue));
  memset(Father,0,sizeof(Father));
  for(int i=1;i<=n;i++)
    Base[i]=i;
  while(!q.empty()) q.pop();
  q.push(Start); InQueue[Start]=true;
  Finish=0;

  while(!q.empty())
    {
      int u=q.front(); q.pop();
      InQueue[u]=false;
      for(int i=1;i<=n;i++)
        {
          if(i==u||G[u][i]==false) continue;
          int v=i;
          if(Base[u]!=Base[v]&&Match[u]!=v)
            {
              if(v==Start||(Match[v]>0&&Father[Match[v]]>0))
                BlossomContract(u,v);
              else if(Father[v]==0)
                {
                  Father[v]=u;
                  if(Match[v]>0)
                    {
                      q.push(Match[v]);
                      InQueue[Match[v]]=true;
                    }
                  else
                    {
                      Finish=v;
                      return ;
                    }
                }
            }
        }
    }
}

void AugmentPath()
{
  int u,v,w;
  u=Finish;
  while(u>0)
    {
      v=Father[u];
      w=Match[v];
      Match[v]=u;
      Match[u]=v;
      u=w;
    }
}

void Edmonds()
{
  memset(Match,0,sizeof(Match));
  for(int u=1;u<=n;u++)
    {
      if(Match[u]==0)
        {
          Start=u;
          FindAugmentingPath();
          if(Finish>0) AugmentPath();
        }
    }
}

int bian[200][2];

int main()
{
  while(scanf("%d%d",&n,&m)!=EOF)
    {
      memset(G,0,sizeof(G));
      memset(TG,0,sizeof(TG));
      ans.clear();

      for(int i=0;i<m;i++)
        {
          int u,v;
          scanf("%d%d",&u,&v);
          bian[i][0]=u;bian[i][1]=v;
          G[u][v]=G[v][u]=1;
          TG[u][v]=TG[v][u]=1;
        }

      Edmonds();

      Count=0;
      for(int i=1;i<=n;i++)
        if(Match[i]) Count++;

      for(int i=0;i<m;i++)
        {
          int u=bian[i][0],v=bian[i][1];
          ///clear about u,v
          for(int j=1;j<=n;j++)
            {
              G[u][j]=G[j][u]=G[j][v]=G[v][j]=0;
            }

          Edmonds();

          int C2=0;
          for(int j=1;j<=n;j++)
              if(Match[j]) C2++;

          if(C2<Count-2)
            ans.push_back(i+1);

          ///Recover
          for(int j=1;j<=n;j++)
            {
              G[u][j]=TG[u][j]; G[j][u]=TG[j][u];
              G[v][j]=TG[v][j]; G[j][v]=TG[j][v];
            }
        }
        int sz = ans.size();
        printf("%d\n",sz);
        for(int i=0;i<sz;i++)
        {
            printf("%d",ans[i]);
            if(i<sz-1)printf(" ");
        }
        printf("\n");
    }
  return 0;
}