Codeforces Round #254 (Div. 2)B. DZY Loves Chemistry

佚名 6年前 (2019-05-07) 随笔 811人围观抢沙发百度已收录

B. DZY Loves Chemistry time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

DZY loves chemistry, and he enjoys mixing chemicals.

SRE实战互联网时代守护先锋，助力企业售后服务体系运筹帷幄！一键直达领取阿里云限量特价优惠。

DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.

Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.

Find the maximum possible danger after pouring all the chemicals one by one in optimal order.

Input

The first line contains two space-separated integers n and m .

Each of the next m lines contains two space-separated integers xi and yi (1 ≤ xi < yi ≤ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.

Consider all the chemicals numbered from 1 to n in some order.

Output

Print a single integer — the maximum possible danger.

Sample test(s) input

1 0

output

input

2 1
1 2

output

input

3 2
1 2
2 3

output

思路：依据反应关系找到全部的集合，再在集合中求得结果（由于n<=50。所以用__int64）

代码：

#include <stdio.h>
int father[55];
int find(int x)
{
	if (father[x] == x)
		return x;
	else
		return (father[x] = find(father[x]));
}
void merge(int a, int b)
{
	int x, y;
	x = find(a);
	y = find(b);
	if (x != y)
		father[x] = y;
}
int main()
{
	int n, m;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		for (int i = 1; i <= n; i++)
			father[i] = i;
		while (m--){
			int x, y;
			scanf("%d%d", &x, &y);
			merge(x, y);
		}
		__int64 ans = 1;
		for (int i = 1; i <= n; i++){
			int fa = father[i];
			if (fa == i){
				int s = 2;
				for (int j = 1; j <= n; j++)
				if (find(j) == fa&&i != j){
					ans = ans*s;
				}
			}
		}
		printf("%I64d\n", ans);
	}
	return 0;
}