hdu 3829 Cat VS Dog 二分图匹配 最大点独立集

Cat VS Dog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)

Problem Description

The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislike-animal, if the child's like-animal is a cat, then his/hers dislike-animal must be a dog, and vice versa.
Now the zoo administrator is removing some animals, if one child's like-animal is not removed and his/hers dislike-animal is removed, he/she will be happy. So the administrator wants to know which animals he should remove to make maximum number of happy children.

 

Input

The input file contains multiple test cases, for each case, the first line contains three integers N <= 100, M <= 100 and P <= 500.
Next P lines, each line contains a child's like-animal and dislike-animal, C for cat and D for dog. (See sample for details)

 

Output

For each case, output a single integer: the maximum number of happy children.

 

Sample Input

1 1 2 C1 D1 D1 C1 1 2 4 C1 D1 C1 D1 C1 D2 D2 C1

 

Sample Output

1 3

Hint

Case 2: Remove D1 and D2, that makes child 1, 2, 3 happy.

 

Source

2011 Multi-University Training Contest 1 - Host by HNU

题意：给你m只狗跟n只猫，p个人，每个人有一只喜欢和不喜欢的动物；

　　　可以选取一些动物，使得满足条件的人最多；

　　　满足条件：一个人的喜欢的动物在，并且不喜欢的动物不在；

思路：二分图匹配最大点独立集模型=所有人-最大匹配；

　　　把所有人的对立的关系连边；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-6
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e3+10,M=1e6+10,inf=1e9+10;
const LL INF=5e17+10,mod=1e9+7;

int n,m;
int mp[N][N];
int linker[N];
bool used[N];
bool dfs(int a)
{
    for(int i=0;i<n;i++)
      if(mp[a][i]&&!used[i])
      {
          used[i]=true;
          if(linker[i]==-1||dfs(linker[i]))
          {
              linker[i]=a;
              return true;
          }
      }
      return false;
}
int hungary()
{
    int result=0;
    memset(linker,-1,sizeof(linker));
    for(int i=0;i<n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i))  result++;
    }
    return result;
}
string l[N],disl[N];
int main()
{
    int c,d;
    while(~scanf("%d%d%d",&c,&d,&n))
    {
        memset(mp,0,sizeof(mp));
        for(int i=0;i<n;i++)
        cin>>l[i]>>disl[i];
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(l[i]==disl[j]||disl[i]==l[j])
                    mp[i][j]=1;
            }
        }
        int cnt=hungary();
        printf("%d
",n-cnt/2);
    }
    return 0;
}