HDU 3829 Cat VS Dog （二分匹配求最大独立集）

Cat VS Dog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1770    Accepted Submission(s): 600

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

 

 

Recommend

xubiao

 

 

此题一看就想到用二分匹配做。但是考虑到可能很多小孩喜欢一样的，感觉二分匹配不行。。。然后又想到最大权匹配。。。结果还是不行。

 

最后转换下思维，对小孩进行二分匹配。就是把小孩当成点。矛盾的小孩间建立一条边。

这样就转化成求最大独立集了。

 

/*

HDU 3829

求最大独立集

*/



#include<stdio.h>

#include<iostream>

#include<algorithm>

#include<string.h>

#include<vector>

using namespace std;



//************************************************

const int MAXN=1505;//这个值要超过两边个数的较大者，因为有linker

int linker[MAXN];

bool used[MAXN];

vector<int>map[MAXN];

int uN;

bool dfs(int u)

{

    for(int i=0;i<map[u].size();i++)

    {

        if(!used[map[u][i]])

        {

            used[map[u][i]]=true;

            if(linker[map[u][i]]==-1||dfs(linker[map[u][i]]))

            {

                linker[map[u][i]]=u;

                return true;

            }

        }

    }

    return false;

}

int hungary()

{

    int u;

    int res=0;

    memset(linker,-1,sizeof(linker));

    for(u=0;u<uN;u++)

    {

        memset(used,false,sizeof(used));

        if(dfs(u)) res++;

    }

    return res;

}

//*****************************************************

char like[MAXN][5];

char dislike[MAXN][5];

int main()

{

    int n,m,p;

    while(scanf("%d%d%d",&n,&m,&p)!=EOF)

    {

        for(int i=0;i<MAXN;i++)map[i].clear();

        for(int i=0;i<p;i++)

        {

            scanf("%s%s",&like[i],&dislike[i]);

        }

        uN=p;

        for(int i=0;i<p;i++)

          for(int j=i+1;j<p;j++)

            if(strcmp(like[i],dislike[j])==0 || strcmp(like[j],dislike[i])==0)

            {

                map[i].push_back(j);

                map[j].push_back(i);

            }



        printf("%d\n",p-hungary()/2);

    }

    return 0;

}