Cat vs. Dog
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1588 Accepted Submission(s): 602
vote on which pets should stay and which should be forced to leave the show.
Each viewer gets to cast a vote on two things: one pet which should be kept on the show, and one pet which should be thrown out. Also, based on the universal fact that everyone is either a cat lover (i.e. a dog hater) or a dog lover (i.e. a cat hater), it has
been decided that each vote must name exactly one cat and exactly one dog.
Ingenious as they are, the producers have decided to use an advancement procedure which guarantees that as many viewers as possible will continue watching the show: the pets that get to stay will be chosen so as to maximize the number of viewers who get both
their opinions satisfied. Write a program to calculate this maximum number of viewers.
* One line with three integers c, d, v (1 ≤ c, d ≤ 100 and 0 ≤ v ≤ 500): the number of cats, dogs, and voters.
* v lines with two pet identifiers each. The first is the pet that this voter wants to keep, the second is the pet that this voter wants to throw out. A pet identifier starts with one of the characters `C' or `D', indicating whether the pet is a cat or dog,
respectively. The remaining part of the identifier is an integer giving the number of the pet (between 1 and c for cats, and between 1 and d for dogs). So for instance, ``D42'' indicates dog number 42.
* One line with the maximum possible number of satisfied voters for the show.
2 1 1 2 C1 D1 D1 C1 1 2 4 C1 D1 C1 D1 C1 D2 D2 C1
1 3
/*
hdoj 2768 最大独立集
题意：有v个观众，每个人投给自己喜欢的猫(或者狗)和讨厌的狗(或者猫)，
如果出现喜欢的和别人讨厌的相同，则其中一人会不满意。
现要求得是最大满意的观众是多少。
方法：把观众分成两个集合，一个是投留下猫的，一个是投留下狗的，
如果存在冲突，则在这两个观众之间加一条边
现在选择最多的顶点，要求各个顶点之间没有线相连，即不出现矛盾。
就是求最大独立集。
1 2 4 左边猫 右边狗
C1 D1 1 C1 4 D2
C1 D1 2 C1
C1 D2 3 C1
D2 C1 上述中3跟4有条边 因为矛盾 所以最大的独立集3个点1 2 3都不相连
最大独立集，在二分图G中，找出点数最多的子图，使得这些点之间都不相连。
最大独立集=顶点数n-最大匹配
*/
#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;
#define N 501
struct node{//记得加 #include<string>
string love,hate;
};
node cat[N],dog[N];
int map[N][N],vis[N],match[N];
int c,d,v,catNum,dogNum;
int dfs(int u)
{
int i;
for(i=0;i<dogNum;i++)
{
if(map[u][i]&&!vis[i])
{
vis[i]=1;
if(match[i]==-1||dfs(match[i]))
{
match[i]=u;
return 1;
}
}
}
return 0;
}
int maxMatch()
{
int i,ans=0;
memset(match,-1,sizeof(match));
for(i=0;i<catNum;i++)
{
memset(vis,0,sizeof(vis));
if(dfs(i))
ans++;
}
return ans;
}
int main()
{
int t,i,j,ans;
char str1[10],str2[10];
//freopen("test.txt","r",stdin);
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&c,&d,&v);
catNum=dogNum=0;
for(i=0;i<v;i++)
{
scanf("%s",str1);
scanf("%s",str2);
if(str1[0]=='C')
{
cat[catNum].love=str1;
cat[catNum++].hate=str2;
}
else
{
dog[dogNum].love=str1;
dog[dogNum++].hate=str2;
}
}
memset(map,0,sizeof(map));
for(i=0;i<catNum;i++)
for(j=0;j<dogNum;j++)
{
if(cat[i].love==dog[j].hate||cat[i].hate==dog[j].love)
map[i][j]=1;
}
ans=maxMatch();
printf("%d\n",v-ans);
}
return 0;
}