cf445B DZY Loves Chemistry

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.

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

1

input

2 1
1 2

output

2

input

3 2
1 2
2 3

output

4

Note

In the first sample, there's only one way to pour, and the danger won't increase.

In the second sample, no matter we pour the 1st chemical first, or pour the 2nd
chemical first, the answer is always 2.

In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).

英文题真是伤不起啊

题意是说给定n个点m条双向边，一开始图是空的，用不同的顺序每次取一个点加入图中，如果图中有和它联通的点，那么ans*=2，最后求max(ans)

实际上对于图中的一个联通块，在联通块中无论加点的顺序如何，它对答案的贡献都是ans*=2^(个数-1)。当然图中有很多联通块，这随便乱搞一下就A了。反正我写的广搜

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m;
long long ans=1;
bool mark[1000];
int map[101][101];
inline long long bfs(int s)
{
	if (mark[s]) return 1;
	int q[10000]={0},t=0,w=1;
	q[1]=s;mark[s]=1;
	long long sigma=1;
	while (t<w)
	{
		int now=q[++t];
		for (int i=1;i<=n;i++)
		  if (!mark[i]&&map[now][i])
		  {
		  	mark[i]=1;
		  	sigma*=2;
		  	q[++w]=i;
		  }
	}
	return sigma;
}
int main()
{
	 n=read();
	 m=read();
	 for (int i=1;i<=m;i++)
	 {
	 	int x=read(),y=read();
	 	map[x][y]=1;
	 	map[y][x]=1;
	 }
	 cout<<endl;
	 for (int i=1;i<=n;i++)
	 {
	 	ans*=bfs(i);
	 }
	printf("%lld",ans);
}