HDU 2120 Ice_cream's world I

最新推荐文章于 2020-02-07 11:05:28 发布

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最新推荐文章于 2020-02-07 11:05:28 发布

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文章标签： php 数据结构与算法

原文链接：http://www.cnblogs.com/zlrrrr/p/9713294.html

本文介绍了一种使用并查集数据结构解决冰激凌女王土地分配问题的方法，通过建立观察塔和墙来分割国家，以确定最多可以奖励多少勤奋的ACMer。算法的时间复杂度为O(m)，并通过实例演示了如何实现。

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

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\n", cnt);
	}
	return 0;
}

转载于:https://www.cnblogs.com/zlrrrr/p/9713294.html