codeforce 445B 並查集

life4711發表於2014-07-07
並查集
http://codeforces.com/problemset/problem/445/B
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).

題目大意：給定一個容器，初始時刻容器的值為1，每放入一個化學藥品若和容器中已有的物品能發生反應則該容器的值乘2，否則的話保持不變。問當所有的化學藥品全部都放入容器完畢後，容器的值為多少。（重複的反應不做累計）

大體思路：

           對於所有能夠反應的化學藥品建立在一課樹上，我們放入物品時只要按照一定的順序，就能保證該樹中沒一個節點放入時都能反應一次。

 這樣題目就轉化為了，n個化學物品能組成多少棵樹k，那麼就反應n-k次。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int root[53];
int get_root(int x)
{
    if(x==root[x])
        return x;
    return root[x]=get_root(root[x]);
}
int main()
{
    int n,m;
    long long ans=1;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            root[i]=i;
        int x,y;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            x=get_root(x);
            y=get_root(y);
            root[x]=y;
        }
        int cnt=n;
        for(int i=1;i<=n;i++)
        {
            if(root[i]==i)
                cnt--;
        }
        for(int i=1;i<=cnt;i++)
                ans*=2;
        printf("%lld\n",ans);
    }
    return 0;
}


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