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.
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)
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.
题目大意：院子里有m条狗，n条猫。p个小孩。这p个小孩，每个小孩要么喜欢狗，讨厌猫；要么喜欢猫，讨厌狗。管理员要把一些狗或者猫驱赶走，如果某个小孩喜欢的动物没被赶走且不喜欢的动物被赶走，他就会高兴。问你最多能让多少小孩高兴。
解题思路：最大独立集：选择尽量多的结点，使得结点之间没有边。喜欢某条狗的小孩会跟不喜欢这条狗的小孩有矛盾，同样猫也一样。在有矛盾的小孩之间连一条边。然后求解最大独立集，即剩下的小孩都没有矛盾。由于不是选择的真正的X部，Y部，而是采用的拆点，连了双向边，所以最后最大匹配应该除以2。
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 1000;
const int INF = 0x3f3f3f3f;
vector<int>G[maxn];
int Mx[maxn], My[maxn], dx[maxn], dy[maxn], used[maxn], dis;
int Map[maxn][maxn];
bool SearchP(int _n){
queue<int>Q;
memset(dx,-1,sizeof(dx));
memset(dy,-1,sizeof(dy));
int dis = INF;
for(int i = 1; i <= _n; i++){
if(Mx[i] == -1){
dx[i] = 0;
Q.push(i);
}
}
int v;
while(!Q.empty()){
int u = Q.front(); Q.pop();
if(dx[u] > dis) break;
for(int i = 0; i < G[u].size(); i++){
v = G[u][i];
if(dy[v] == -1){
dy[v] = dx[u] + 1;
if(My[v] == -1){
dis = dy[v];
}else{
dx[My[v]] = dy[v] + 1;
Q.push(My[v]);
}
}
}
}
return dis != INF;
}
int dfs(int u){
int v;
for(int i = 0; i < G[u].size(); i++){
v = G[u][i];
if(!used[v] && dy[v] == dx[u] + 1){
used[v] = 1;
if(My[v] != -1 && dy[v] == dis){
continue;
}
if(My[v] == -1 || dfs(My[v])){
Mx[u] = v;
My[v] = u;
return true;
}
}
}
return false;
}
int MaxMatch(int ln,int rn){
int ret = 0;
memset(Mx,-1,sizeof(Mx));
memset(My,-1,sizeof(My));
while(SearchP(ln)){
memset(used,0,sizeof(used));
for(int i = 1; i <= ln; i++){
if(Mx[i] == -1 && dfs(i)){
ret++;
}
}
}
return ret;
}
char like[maxn][20], dislike[maxn][20];
int main(){
int T, cas = 0, n, m, N, M, k, P;
while(scanf("%d%d%d",&N,&M,&P)!=EOF){
for(int i = 0; i <= P; i++){
G[i].clear();
}
for(int i = 1; i <= P; i++){
scanf("%s%s",like[i],dislike[i]);
}
for(int i = 1; i <= P; i++){
for(int j = i+1; j <= P; j++){
if(strcmp(like[i],dislike[j])== 0 || strcmp(dislike[i],like[j]) == 0){
G[i].push_back(j);
G[j].push_back(i);
}
}
}
n = m = P;
int res = MaxMatch(n,m);
printf("%d
", n - res/2);
}
return 0;
}