HDU 2120 Ice_cream's world I

http://acm.hdu.edu.cn/showproblem.php?pid=2120

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

 

题解：并查集查有多少个圈

时间复杂度：$O(m)$

代码：

#include <bits/stdc++.h>
using namespace std;

int f[10010];
int n, m, cnt;

void init() {
	for(int i = 0; i < 10010; i ++)
	f[i] = i;
}

int Find(int x) {
  if(x != f[x]) f[x] = Find(f[x]);
  return f[x];
}

void Merge(int x, int y) {
	int fx = Find(x);
	int fy = Find(y);
	if(fy != fx)
        f[fy] = fx;
	else
        cnt++;
}

int main() {
	while(~scanf("%d%d", &n, &m)) {
		init();
		int a, b;
		cnt = 0;
		for(int i = 0; i < m; i ++) {
			scanf("%d%d", &a, &b);
			Merge(a, b);
		}
		printf("%d
", cnt);
	}
	return 0;
}