HDU3829_Cat VS Dog

题目是这样的，给定一些人喜欢某只猫或者狗，讨厌某只猫或者狗。求最多能够同时满足多少人的愿望？

题目很有意思。建模后就很简单了。

对于同一只猫或者狗，如果有一个讨厌，另一个人喜欢，那么这两个连一条边。最终，最大独立集数等于最大匹配数就可以了。。

Orz。

召唤代码君：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#define maxn 505
using namespace std;

vector<int> likec[maxn],disc[maxn],liked[maxn],disd[maxn];
vector<int> v[maxn];
int n,m,p,ans,t1,t2;
int f[maxn],tag[maxn];
char s1[maxn],s2[maxn];

void init()
{
	ans=0;
	for (int i=1; i<=p; i++) v[i].clear(),f[i]=0,tag[i]=0;
	for (int i=1; i<=n; i++) likec[i].clear(),disc[i].clear();
	for (int i=1; i<=m; i++) liked[i].clear(),disd[i].clear();
}

int num(char S[])
{
	int cur=0;
	for (int i=0; S[i]; i++) cur=cur*10+S[i]-'0';
	return cur;
}

int dfs(int cur,int T)
{
	if (tag[cur]==T) return false;
		else tag[cur]=T;
	for (unsigned i=0; i<v[cur].size(); i++)
	{
		if (tag[v[cur][i]]==T) continue;
		if (f[v[cur][i]]==0 || dfs(f[v[cur][i]],T))
		{
			f[v[cur][i]]=cur;
			f[cur]=v[cur][i];
			return true;
		}
	}
	return false;
}

int main()
{
	while (scanf("%d%d%d",&n,&m,&p)!=EOF)
	{
		init();
		for (int i=1; i<=p; i++)
		{
			scanf("%s%s",s1,s2);
			t1=num(s1+1),t2=num(s2+1);
			if (s1[0]=='C') likec[t1].push_back(i);
				else liked[t1].push_back(i);
			if (s2[0]=='C') disc[t2].push_back(i);
				else disd[t2].push_back(i);
		}
		for (int i=1; i<=n; i++)
			for (unsigned x1=0; x1<likec[i].size(); x1++)
				for (unsigned x2=0; x2<disc[i].size(); x2++)
				{
					v[likec[i][x1]].push_back(disc[i][x2]);
					v[disc[i][x2]].push_back(likec[i][x1]);
				}
		for (int i=1; i<=m; i++)
			for (unsigned x1=0; x1<liked[i].size(); x1++)
				for (unsigned x2=0; x2<disd[i].size(); x2++)
				{
					v[liked[i][x1]].push_back(disd[i][x2]);
					v[disd[i][x2]].push_back(liked[i][x1]);
				}
		for (int i=1; i<=p; i++)
		{
			if (f[i]!=0) continue;
			if (dfs(i,i)) ans++;
		}
		printf("%d\n",p-ans);
	}
	return 0;
}