Codeforces987F

题目链接：

http://codeforces.com/contest/987/problem/F

F. AND Graph

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of size $m$

with integer elements between $0$ and $2^n-1$ inclusive. Let's build an undirected graph on these integers in the following way: connect two integers $x$ and $y$ with an edge if and only if $x\&y=0$. Here $\&$

is the bitwise AND operation. Count the number of connected components in that graph.

Input

In the first line of input there are two integers $n$

and $m$ ( $0\le n\le 22$, $1\le m\le 2^n$

).

In the second line there are $m$

integers $a_1,a_2,\dots ,a_m$ ( $0\le a_i<2^n$) — the elements of the set. All $a_i$

are distinct.

Output

Print the number of connected components.

Examples

Input

Copy

2 3
1 2 3

Output

Copy

2

Input

Copy

5 5
5 19 10 20 12

Output

Copy

2

Note

Graph from first sample:

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Graph from second sample:

题意:给你一个n，一个m，m < 2^n， 然后就是如果 x & y == 0 的话就可以将这2个数连起来， 求最后的有几块。

题解: 不难想到，x有边连出的一定是 (2^n-1) ^ x 的一个子集，直接连子集复杂度是爆炸的。。。但是我们可以一个1一个1的消去，最后变成补集的一个子集。

但是必须当且仅当 至少有一个 a 等于 x 的时候， 可以直接dfs(all ^ x) ,否则直接消1连边。。。

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1<<22;

int n, m, a[maxn], all;
int vis[maxn];
int have[maxn];

void dfs(int x)
{
    if(vis[x]) return;
    vis[x] = 1;
    if(have[x]) dfs(all ^ x);
    for(int i = 0;i < n;i ++) {
        if(x & 1 << i) dfs(x ^ 1 << i);
    }
}
int main()
{
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> n >> m;
    for(int i = 1;i <= m;i ++) cin >> a[i];
    for(int i = 1;i <= m;i ++) have[a[i]] = 1;
    all = 1<<n;
    all--;
    int res = 0;
    for(int i = 1;i <= m;i ++) {
        if(!vis[a[i]]) {
            vis[a[i]] = 1;
            res ++;
            dfs(all ^ a[i]);
        }
    }
    cout << res << '\n';
    return 0;
}