hdoj--2120--Ice_cream's world I（并查集判断环）

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 975    Accepted Submission(s): 567

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition
the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

 

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between
A and B has a wall(A and B are distinct). Terminate by end of file.

 

Output

Output the maximum number of ACMers who will be awarded.

One answer one line.

 

Sample Input

8 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7

 

Sample Output

3

 

Author

Wiskey

 

Source

HDU 2007-10 Programming Contest_WarmUp

 

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好久不写并查集，刚开始以为会是scc什麽的，还好没写- -||

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int pre[10010],n,m,cnt;
void init()
{
	for(int i=0;i<10010;i++)
	pre[i]=i;
}
int find(int x)
{
	return pre[x]==x?x:find(pre[x]);
}
void join(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fy!=fx)
	pre[fy]=fx;
	else
	cnt++;
}
int main()
{
	while(cin>>n>>m)
	{
		init();
		int a,b;
		cnt=0;
		for(int i=0;i<m;i++)
		{
			cin>>a>>b;
			join(a,b);
		}
		cout<<cnt<<endl;
	}
	return 0;
}