hdu3829（二分图，最大独立集）

Cat VS Dog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 3326    Accepted Submission(s): 1161

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. 
    
   
   
   
   

重点不在二分图，重点是怎样看出来这个题用二分图写，或者，怎样去建图，怎样去匹配。。。。

题目只是给出了两个集合，猫，狗，还有剩下一群人，可是建图的过程却是建立在人上面的。。。。

需要注意的是这一题不是喜欢狗就是喜欢猫，所以是个非常明显的二分图，所以是建立在二分图上的最大独立集，所以用最大独立集数=顶点数-最大匹配数。如果构不成二分图，是不能用最大匹配来解决的，求非二分图的最大独立集相当于求其补图最大团。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
const int N=555;
bool tu[N][N];
int from[N];///记录右边的点如果配对好了它来自哪里
bool use[N];///记录右边的点是否已经完成了配对
int n,m;///m,n分别表示两边的各自数量,n是左边，m是右边
bool dfs(int x)
{
    for(int i=1;i<=m;i++)///m是右边，所以这里上界是m
    if(!use[i]&&tu[x][i])
    {
        use[i]=1;
        if(from[i]==-1||dfs(from[i]))
        {
            from[i]=x;
            return 1;
        }
    }
    return 0;
}
int hungary()
{
    int tot=0;
    memset(from,-1,sizeof(from));
    for(int i=1;i<=n;i++)///n是左边，所以这里上界是n
    {
        memset(use,0,sizeof(use));
        if(dfs(i))
            tot++;
    }
    return tot;
}
char x1[505],x2[505];int t1[505],t2[505];
int main()
{
    int M,N,p;
    while(cin>>N>>M>>p)
    {
        m=n=p;
        memset(tu,0,sizeof(tu));
        for(int i=1;i<=p;i++)
        {
            scanf(" %c%d %c%d",&x1[i],&t1[i],&x2[i],&t2[i]);
            for(int j=0;j<i;j++)
            if(x1[j]==x2[i]&&t1[j]==t2[i])
            tu[i][j]=tu[j][i]=1;
            else if(x2[j]==x1[i]&&t2[j]==t1[i])
            tu[i][j]=tu[j][i]=1;
        }
        printf("%d\n",p-hungary()/2);
    }
    return 0;
}