HDU3829：Cat VS Dog（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): 4751 Accepted Submission(s): 1757

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 D2

D2 C1

Sample Output:

1
3

Hint:

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

题意：

每个人都会喜欢一只猫（狗），相应的他会讨厌一只狗（猫），问现在要去掉多少猫or狗，可以让最多的人满意。

题解：

我一开始想的是对人和动物建图，让一个人快乐的话，就是留住一个动物以及去掉一个动物，但这样有点复杂，没做出来...

然后看了下其它的题解，发现如果一个人喜欢一只猫，另外一个人讨厌一只猫的话，那么就不可能让这两个人同时满意，最多只能让一个人满意。

这样问题就可以转化为最大独立集了，最大独立集的意思就是选取尽量多的点，并且这些点互不相邻。

想到这里建图就比较好建了，对人人建图。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define mem(x) memset(x,0,sizeof(x))
using namespace std;

const int N = 505;
int n,m,p,ans;
char like[N][5],dlike[N][5]; //这里我一开始数组开的是4，WA了，开成5就AC了 
int check[N],match[N],link[N][N];

inline int dfs(int x){
    for(int i=1;i<=p;i++){
        if(link[x][i] && !check[i]){
            check[i]=1;
            if(!match[i] || dfs(match[i])){
                match[i]=x;
                return 1;
            }
        }
    }
    return 0;
}

int main(){
    while(~scanf("%d%d%d",&n,&m,&p)){
        mem(match);mem(link);ans=0;
        for(int i=1;i<=p;i++){
            scanf("%s%s",like[i],dlike[i]);
        }
        for(int i=1;i<=p;i++)
            for(int j=1;j<=p;j++)
                if(strcmp(like[i],dlike[j])==0 || strcmp(like[j],dlike[i])==0) link[i][j]=1;
        for(int i=1;i<=p;i++){
            mem(check);
            if(dfs(i)) ans++;
        }
        printf("%d
",p-ans/2);
    }
    return 0;
}