标签：first include second ant rect ISE ati return int
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。。。
Discription
You are given a set of size mm with integer elements between 00 and 2n?12n?1 inclusive. Let‘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 nn and mm (0≤n≤220≤n≤22, 1≤m≤2n1≤m≤2n).
In the second line there are mm integers a1,a2,…,ama1,a2,…,am (0≤ai<2n0≤ai<2n) — the elements of the set. All aiai are distinct.
Output
Print the number of connected components.
Examples
2 3
1 2 3
2
5 5
5 19 10 20 12
2
Note
Graph from first sample:
Graph from second sample:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=5000005;
int ci[233],T,n,a[maxn],ans,all;
bool v[maxn],isp[maxn];
inline int read(){
int x=0; char ch=getchar();
for(;!isdigit(ch);ch=getchar());
for(;isdigit(ch);ch=getchar()) x=x*10+ch-‘0‘;
return x;
}
void dfs(int x){
if(v[x]) return;
v[x]=1;
if(isp[x]) dfs(all^x);
for(int i=0;i<=T;i++) if(ci[i]&x) dfs(x^ci[i]);
}
inline void solve(){
for(int i=1;i<=n;i++) if(!v[a[i]]){
ans++,v[a[i]]=1,dfs(all^a[i]);
}
}
int main(){
ci[0]=1;
for(int i=1;i<=22;i++) ci[i]=ci[i-1]<<1;
T=read(),n=read(),all=ci[T]-1;
for(int i=1;i<=n;i++) a[i]=read(),isp[a[i]]=1;
solve();
printf("%d\n",ans);
return 0;
}
标签：first include second ant rect ISE ati return int
原文地址：https://www.cnblogs.com/JYYHH/p/9112798.html
