HDU 3829 Cat VS Dog 最大独立集(最大匹配)

Cat VS Dog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 3570    Accepted Submission(s): 1259

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 2C1 D1D1 C11 2 4C1 D1C1 D1C1 D2D2 C1

 

Sample Output

1
3
思路：所有喜欢狗的孩子划到左边集合，喜欢猫的孩子划到右边集合，这样构成一个二分图。当左边的孩子喜欢的狗和右边的讨厌的狗相同或左边的孩子讨厌的猫和右边的孩子喜欢的猫相同时建边。这样问题转化为最大独立集。
最大独立集=节点数-最大匹配数。
#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;typedef struct{int c,like,dlike;}H;H l[550],r[550];int map[550][550],vis[550],flag[550],nx,ny;int find(int k);int main(){int n,i,j,k,sum,m,p,x,y;char a,b;while(scanf("%d%d%d",&n,&m,&p)!=EOF){nx=1;ny=1;memset(map,0,sizeof(map));memset(flag,0,sizeof(flag));for(i=1;i<=p;i++){scanf(" %c%d %c%d",&a,&x,&b,&y);if(a=='D'){l[nx].like=x;l[nx].dlike=y;l[nx++].c=i;}else{r[ny].like=x;r[ny].dlike=y;r[ny++].c=i;}}for(i=1;i<nx;i++){for(j=1;j<ny;j++){if(l[i].like==r[j].dlike)map[l[i].c][r[j].c]=1;if(l[i].dlike==r[j].like)map[l[i].c][r[j].c]=1;}}sum=0;for(i=1;i<nx;i++){memset(vis,0,sizeof(vis));if(find(l[i].c))sum++;}printf("%d\n",p-sum);}return 0;}int find(int k){int i,j,u,v;for(j=1;j<ny;j++){u=k;v=r[j].c;//printf("%d %d\n",u,v);if(!vis[v] && map[u][v]){vis[v]=1;//printf("%d %d\n",u,v);if(!flag[v] || find(flag[v])){flag[v]=u;return 1;}}}return 0;}