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

链接：http://codeforces.com/contest/445/problem/B

题目：

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 by2. 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 andm


.

Each of the next m lines contains two space-separated integersxi andyi(1 ≤ xi < yi ≤ n).
These integers mean that the chemicalxi will react with the chemicalyi. 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 11 2

Output

2

Input

3 21 22 3

Output

4

解题思路;

并查集，统计共有几个集合，对于每个集合，最佳的情况（发生化学反应的次数）是集合中的元素个数-1，统计一共发生多少次化学反应。

代码：

#include <cstdio>
#include <cstring>
using namespace std;

const int MAXN = 55;
int n, m, set[MAXN], vis[MAXN];

int set_find(int p)
{
    if(set[p] < 0) return p;
    return set[p] = set_find(set[p]);
}

void join(int p, int q)
{
    p = set_find(p);
    q = set_find(q);
    if(p != q) set[p] = q;
}

int main()
{
    int ans = 0;
    memset(set, -1, sizeof(set));
    memset(vis, 0, sizeof(vis));
    scanf("%d%d", &n, &m);
    for(int i = 0; i < m; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        join(x, y);
    }
    for(int i = 1; i <= n; i++)
    {
        if(vis[i]) continue;
        int t = 0, p = i;
        while(set[p] > 0)
        {
            vis[p] = 1;
            p = set[p];
            if(!vis[p]) t++;
        }
        ans += t;
    }
    long long ret = 1;
    for(int i = 0; i < ans; i++)
    {
        ret = ret * 2;
    }
    printf("%I64d\n", ret);
    return 0;
}