CodeForces

CodeForces - 445B 贪心+模拟

CodeForces - 445B

B. DZY Loves Chemistry
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

Examples

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).

题目大意：有n种化学物质，其中有m对化学物质两两互相反应。现在要把他们按照某种按照顺序加入试管，求危险程序最大是多少。(空试管的危险度为1，每次加入一种化学物质，如果能够发生反应，那么危险程度翻倍。)

这道题目居然是用并查集来做，我居然用了贪心+模拟。。。。

#include <bits/stdc++.h>using namespace std;#define LL long longstruct node{int x,y;node(){x = 0,y = 0;}};bool cmp(node a,node b){    return a.y>b.y;}int main(){    int n,m,a,b;    while(~scanf("%d%d",&n,&m))    {        node f[n+1];        int vis[n+1] = {0};        queue<int> qu;        vector<int> v[n+1];        for(int i = 1;i <= m;i++)        {            scanf("%d%d",&a,&b);            v[a].push_back(b);            v[b].push_back(a);            f[a].x = a,f[a].y++;            f[b].x = b,f[b].y++;        }        LL sum = 1;        sort(f+1,f+n+1,cmp);        for(int i = 1;i <= n;i++)        {            if(vis[f[i].x] == 0)            {                qu.push(f[i].x);                vis[f[i].x] = 1;                while(!qu.empty())                {                    int cur = qu.front();                    //cout<<cur<<" **"<<endl;                    qu.pop();                    for(int j = 0;j < v[cur].size();j++)                    {                        int u = v[cur][j];                        if(vis[u] == 0)                        {                            sum *= 2;                            qu.push(u);                            vis[u] = 1;                        }                    }                }            }        }        printf("%lld\n",sum);    }    return 0;}