Hdu 3829 Cat VS Dog

最新推荐文章于 2020-06-20 21:03:16 发布

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最新推荐文章于 2020-06-20 21:03:16 发布

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分类专栏：匹配问题 hdu

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本文探讨了一种算法解决动物园管理员如何通过移除部分动物来最大化孩子们的幸福感的问题。管理员需在N只猫和M只狗中决定哪些动物的移除能让最多的孩子感到快乐，每个孩子喜欢猫或狗之一，讨厌另一种动物。文章提供了具体的输入输出样例和C++代码实现。

题目：

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

代码如下：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
const int maxn=505;
int n,m,p;
int a[maxn][maxn];
int match[maxn];
char dli[maxn][5];
char li[maxn][5];
int vis[maxn];
bool Find(int x)
{
    for (int i=0;i<p;i++)
    {
        if(!vis[i]&&a[x][i])
        {
            vis[i]=1;
            if(match[i]==-1||Find(match[i]))
            {
                match[i]=x;
                return true;
            }
        }
    }
    return false;
}
int algor ()
{
    memset (match,-1,sizeof(match));
    int ans=0;
    for (int i=0;i<p;i++)
    {
        memset (vis,0,sizeof(vis));
        if(Find(i)) ans++;
    }
    return ans;
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&p)!=EOF)
    {
        memset (a,0,sizeof(a));
        for (int i=0;i<p;i++)
        {
            scanf("%s %s",li[i],dli[i]);
            for (int j=0;j<i;j++)
            {
                if(strcmp(li[i],dli[j])==0||strcmp(dli[i],li[j])==0)
                {
                    a[i][j]=1;
                    a[j][i]=1;
                }
            }
        }
        printf("%d\n",p-algor()/2);
    }
    return 0;
}