Codeforces Round #485 (Div. 1) C. AND Graph

C. 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 $\mathrm{mm with\; integer\; elements\; between 00 and 2n-12n-1 inclusive.\; Let\text{'}s\; build\; an\; undirected\; graph\; on\; these\; integers\; in\; the\; following\; way:\; connect\; two\; integers xx and yy with\; an\; edge\; if\; and\; only\; if x\&y=0x\&y=0.\; Here \&\& is\; the bitwise\; AND\; operation.\; Count\; the\; number\; of\; connected\; components\; in\; that\; graph.}$

Input

In the first line of input there are two integers $\mathrm{nn and mm (0\le n\le 220\le n\le 22, 1\le m\le 2n1\le m\le 2n).}$

In the second line there are $\mathrm{mm integers a1,a2,\dots ,ama1,a2,\dots ,am (0\le ai<2n0\le ai<2n) —\; the\; elements\; of\; the\; set.\; All aiai are\; distinct.}$

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:

Graph from second sample:

思路：对于每个x（0 <= x < 1 << n），向刚好从x中去掉某个2进制位的1的所有数y连一条边。然后对于每个a[i]，从~a[i]跑dfs，中途遇到a[j]时，再继续从~a[j]跑，跑完一套dfs就求出了一个联通块。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

int n, m, a[5000015], c;
bool vis[9000015];
int nex[9000015];
void dfs(int x) {
    vis[x] = 1;
    if (~nex[x] && !vis[nex[x]]) dfs(nex[x]);
    if (c <= x) {
        x -= c;
        int p = x;
        for (; p; p -= p & -p) {
            int y = x - (p & -p) + c;
            if (vis[y]) continue;
            dfs(y);
        }
    }
}
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) scanf("%d", &a[i]);
    c = 1 << n;
    clr(nex, -1);
    for (int i = 1; i <= m; ++i) {
        nex[a[i]] = 2 * c - 1 - a[i];
        nex[a[i] + c] = a[i];
    }
    int cnt = 0;
    for (int i = 1; i <= m; ++i) {
        if (!vis[a[i]]) {
            ++cnt;
            dfs(a[i]);
        }
    }
    return !printf("%d
", cnt);
}