HDU 3829 Cat VS Dog ( 最大独立集 = 顶点数 - 最大匹配数)

Cat VS Dog

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2148 Accepted Submission(s): 748

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 D1
C1 D2
D2 C1

Sample Output

1
3

题意：动物园有 N只猫，M只狗，有P个小孩去动物园，现在动物管理员，移除小孩不喜欢的而动物，最多可以使多少个小孩高兴。

思路：建立完二分图之后，相当于求二分图的最大独立集 = 顶点数 - 最大匹配数。

import java.io.*;
import java.util.*;
public class Main {
	int n,m,p;
	int[][] map;
	int[] link=new int[600];
	boolean[] mark=new boolean[600];
	public static void main(String[] args) {
		new Main().work();
	}
	void work(){
		Scanner sc=new Scanner(new BufferedInputStream(System.in));
		while(sc.hasNext()){
			n=sc.nextInt();
			m=sc.nextInt();
			p=sc.nextInt();
			Node node[]=new Node[p];
			map=new int[600][600];
			for(int i=0;i<p;i++){
				String like=sc.next();
				String dis=sc.next();
				node[i]=new Node(like,dis);
			}
			//建立二分图
			for(int i=0;i<p;i++){
				for(int j=i+1;j<p;j++){
					if(node[i].like.equals(node[j].dis)||node[i].dis.equals(node[j].like)){
						map[i][j]=1;
						map[j][i]=1;
					}
				}
			}
			hungary();
		}
	}
	//匈牙利算法
	void hungary(){
		Arrays.fill(link,0);
		int ans=0;
		for(int i=0;i<p;i++){
			Arrays.fill(mark,false);
			if(DFS(i))
				ans++;
		}
		System.out.println(p-ans/2);//每个点用了两次，所以要除以2
	}
	
	
	boolean DFS(int x){
		for(int i=0;i<p;i++){
			if(map[x][i]==1&&!mark[i]){
				mark[i]=true;
				if(link[i]==0||DFS(link[i])){
					link[i]=x;
					return true;
				}
			}
		}
		return false;
	}
	class Node{
		String like;
		String dis;
		Node(String like,String dis){
			this.like=like;
			this.dis=dis;
		}
	}
}

Re: long71751380昨天 19:26: 我刚才看了你给的网站，恩，不错，大家一起学吧

HDU 3829 Cat VS Dog ( 最大独力集 = 顶点数 - 最大匹配数)

Cat VS Dog

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