Cat VS Dog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 4539    Accepted Submission(s): 1670

 

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

 

 

       每个人有喜欢的又不喜欢的，当他喜欢的留下，厌恶的走开始他会开心，问最多使多少人同时开心。

       这样的话，应该对人进行二分，将喜欢及厌恶的两个人连表，求图的最大独立集就对了；

#include<cstdio>
#include<iostream>
#include<vector>
#include<cstring>
#include<string>
using namespace std;
vector<int>G[505];
bool vis[505];
int use[505];
struct ***
{
	string love, hate;
}spt[505];
int n, m, p;
int find(int x) {
	int sz = G[x].size();
	for (int s = 0; s < sz; s++) {
		int v = G[x][s];
		if (!vis[v]) {
			vis[v] = 1;
			if (use[v] == 0 || find(use[v])) {
				use[v] = x;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	ios::sync_with_stdio(0);
	while (cin >> n >> m >> p){
		memset(use, 0, sizeof(use));
		for (int s = 0; s <= p; s++) {
			G[s].clear();
		}
		for (int s = 0; s < p; s++) {
			cin >> spt[s].love >> spt[s].hate;
		}
		for (int s = 0; s < p; s++) {
			for (int w = 0; w < p; w++) {
				if (spt[s].hate == spt[w].love) {
					G[s].push_back(w);
					G[w].push_back(s);
				}
			}
		}
		int sum = 0;
		for (int s = 0; s <= p; s++) {
			memset(vis, 0, sizeof(vis));
			if (find(s))sum++;
		}
		cout << p - (sum / 2) << '\n';
	}
}