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  • codeforces CF986C AND Graph 建圖 dfs

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    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 $ 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:

    pic

    Graph from second sample:

    pic

     

    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
    */
    
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  • 原文地址:https://www.cnblogs.com/PotremZ/p/9600617.html
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