鏈接

大意:給定$m$個數, 若$x\&y=0$, 則在$x$與$y$之間連一條無向邊. 求無向圖的連通塊個數

暴力連邊顯然超時的, 可以通過輔助結點優化連邊, 復雜度$O(n2^n)$

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define pb push_back
using namespace std;

const int N = 1<<22;

int n, m;
int a[N], vis[N], f[N];

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,m) scanf("%d",a+i),f[a[i]]=1;
	int mx = (1<<n)-1, ans = 0;
	REP(i,1,m) if (!vis[a[i]]) {
		++ans;
		queue<int> q;
		q.push(a[i]);
		while (q.size()) {
			int x = q.front();q.pop();
			if (vis[x]) continue;
			vis[x] = 1;
			if (f[x]) q.push(mx^x);
			for (int y=x; y; y^=y&-y) q.push(x^y&-y);
		}
	}
	printf("%d\n", ans);
}

AND Graph CodeForces - 987F (狀壓)