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
HintCase 2: Remove D1 and D2, that makes child 1, 2, 3 happy.
//
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int N=600;//N不能太大 否则超时
int cap[N][N];//初始化要清零
int _link[N];
bool used[N];
int nx,ny;//1->nx
bool _find(int t)
{
for(int i=1;i<=ny;i++)
if(!used[i]&&cap[t][i]==1)
{
used[i]=true;
if(_link[i]==-1||_find(_link[i]))
{
_link[i]=t;
return true;
}
}
return false;
}
int MaxMatch()
{
int num=0;
memset(_link,-1,sizeof(_link));
for(int i=1;i<=nx;i++)
{
memset(used,false,sizeof(used));
if(_find(i)) num++;
}
return num;
}
string a[600],b[600];
//最大点独立集=P-最大匹配
//只有喜欢猫和喜欢狗的人会发生冲突,有用的总人数为p
//如果两边发生冲突，则连一条边
//这样求的是最多不发生冲突的人个数,即最大点独立集
int main()
{
int n,m,p;
while(scanf("%d%d%d",&n,&m,&p)==3)
{
for(int i=1;i<=p;i++)
{
cin>>a[i]>>b[i];
}
memset(cap,0,sizeof(cap));
nx=ny=p;
for(int i=1;i<=p;i++)
{
if(a[i][0]=='C')//喜欢猫的人
for(int j=1;j<=p;j++)
{
if(a[j][0]=='D')//喜欢狗的人
if(a[i]==b[j]||a[j]==b[i])//如果发生冲突
{
cap[i][j]=1;
}
}
}
int ans=MaxMatch();
printf("%d\n",p-ans);
}
return 0;
}
