hdu3829——Cat VS Dog（最大独立集）-白红宇

hdu3829——Cat VS Dog（最大独立集）

阅读量：2343 次

发布时间：2019-05-10

本文共 1497 字，大约阅读时间需要 4 分钟。

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 D2

D2 C1

Sample Output

Hint

Case 2: Remove D1 and D2, that makes child 1, 2, 3 happy.

对小孩拆点，能建边的条件是两个小孩喜欢的动物和不喜欢的正好矛盾，说明这两个小孩需要作出取舍，接下来最大幸福度就是作出最大限度不用作出取舍的小孩的幸福度，其实就是最大独立集

当然以上都是我口胡的

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           //#include 
           #include 
            
             #define INF 0x3f3f3f3f#define MAXN 1505#define Mod 10001using namespace std;bool vis[MAXN];int pre[MAXN],n; //匹配路径;vector
             
               map[MAXN];string like[MAXN],dislike[MAXN];int find(int cur){ for(int i=0; i
              
               >a>>b; like[i]=a; dislike[i]=b; } for(int i=0; i

转载地址：http://ovcvb.baihongyu.com/

你可能感兴趣的文章