codeforces CF986C AND Graph 建圖 dfs
C. AND Graph
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 \le 2^n ) $ — the elements of the set.
All $ a_i $ are distinct.
Output
Print the number of connected components.
Examples
input1
2 3
1 2 3
output1
2
input2
5 5
5 19 10 20 12
output2
2
Note
Graph from first sample:
Graph from second sample:
題目大意
-
給定 $ m $ 個 $ 0 ~ 2^n-1 $ 之間的整數,每個整數代表一個點
-
兩個整數 $ x,y $ 之間有無向邊當且僅當 $ x $ & $ y=0 $ ,求無向圖有多少個連通塊
-
$ n \le 22 $
題解
-
把 $ 0 ~ 2^n-1 $ 之間的每個點拆成 $ x $ 和 $ x' $ 兩個點
-
$ 1. $ 從 $ x $ 到 $ (~x)' $ 連有向邊
-
$ 2. $ 從 $ x' $ 到 $ (x \quad xor \quad (1 \ll k ))' $ 連有向邊 $ (o \le k < n ) $
-
$ 3. $ 若 $ x $ 屬於給定的 $ m $ 個數,則從 $ x' $ 到 $ x $ 連有向邊
-
從 $ m $ 個數出發進行遍歷,求連通塊數
代碼
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn (1<<23)+5
int n,m,ans,tot;
bool vis[maxn],mark[maxn];
void dfs(int u){
if(vis[u]) return; vis[u]=1;
if(u<(1<<n)) dfs(u+(1<<n));
else {
tot=(1<<(n+1))-1-u;
if(!vis[tot]&&mark[tot]) dfs(tot);
for(int i=0;i<n;++i) if(!vis[u|(1<<i)]) dfs(u|(1<<i));
}
}
int main(){
scanf("%d %d",&n,&m);
for(int x,i=1;i<=m;++i){ scanf("%d",&x); mark[x]=1; }
for(int i=0;i<(1<<n);++i)
if(mark[i]&&!vis[i]){ ++ans; dfs(i); }
printf("%d",ans);
return 0;
}
/*
# 40059473
When 2018-07-07 14:30:49
Who PotremZ
Problem C - AND Graph
Lang GNU C++
Verdict Accepted
Time 764 ms
Memory 91100 KB
*/