independent set 1

 independent set 1

時間限制：C/C++ 1秒，其餘語言2秒
空間限制：C/C++ 102400K，其餘語言204800K
64bit IO Format: %lld

題目描述

Note:  For C++ languages, the memory limit is 100 MB. For other languages, the memory limit is 200 MB.

In graph theory, an independent set is a set of nonadjacent vertices in a graph. Moreover, an independent set is maximum if it has the maximum cardinality (in this case, the number of vertices) across all independent sets in a graph.

An induced subgraph G'(V', E') of a graph G(V, E) is a graph that satisfies:

* V′⊆V

* edge (a,b)∈E′ if and only if a∈V′,b∈V′, and edge (a,b)∈E;


Now, given an undirected unweighted graph consisting of n vertices and m edges. This problem is about the cardinality of the maximum independent set of each of the 2^n possible induced subgraphs of the given graph. Please calculate the sum of the 2^n such cardinalities.

輸入描述:

The first line contains two integers n and m (2≤n≤26,0≤m≤n×(n−1)​) --- the number of vertices and the number of edges, respectively. Next m lines describe edges: the i-th line contains two integers xi,yi (0≤xi<yi<n) --- the indices (numbered from 0 to n - 1) of vertices connected by the i-th edge.

The graph does not have any self-loops or multiple edges.

輸出描述:

Print one line, containing one integer represents the answer.

輸入

3 2
0 1
0 2

輸出

9

說明

The cardinalities of the maximum independent set of every subset of vertices are: {}: 0, {0}: 1, {1}: 1, {2}: 1, {0, 1}: 1, {0, 2}: 1, {1, 2}: 2, {0, 1, 2}: 2. So the sum of them are 9.
連接：https://ac.nowcoder.com/acm/contest/885/E
來源：牛客網

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題意：求一個圖的2^n種子圖的最大點獨立集。
思路：

 •咱們能夠用一個 n-bit 2 進制整數來表示一個點集，第 i 個 bit 是 1 就表明點集包含第 i 個 點，如果 0 則不包含 
 • 每一個點相鄰的點也能夠用一個 n-bit 2 進制整數表示，計作 ci，若第 i 個點和第 j 個點相鄰， ci 的第 j 個 bit 是 1，不然是 0
 • 記 x 的最低位且是 1 的 bit 的位置是 lbx
 • 令 dp[x] 表明點集 x 的最大獨立集 size，那麼咱們可以根據點 lbx 是否屬於最大獨立集來列 出如下關係式：
 dp[x] = max(dp[x - (1<<lbx)], dp[x & (~clb_x)] + 1) (使用 c 語言運算符)

 •高效位運算參考博客：https://blog.csdn.net/yuer158462008/article/details/46383635

#include<bits/stdc++.h>
using namespace std;
char dp[1<<26];
int Map[26]={0};
int max(char a,int b)
{
    if(a>b)return a;
    return b;
}
int main()
{    
    int n,m;
    scanf("%d %d",&n,&m);
    
    while(m--)
    {
        int u,v;
        scanf("%d %d",&u,&v);
        Map[u]|=(1<<v);
        Map[v]|=(1<<u);
    }
    for(int i=0;i<n;i++)Map[i]|=1<<i;
    
    int upper=1<<n;
    long long ans=0;
    for(int i=1;i<upper;i++)
    {
        int lbx=__builtin_ctz(i);
        dp[i] = max(dp[i - (1<<lbx)] , dp[i & (~Map[lbx])] + 1);
        ans+=dp[i];
    }
    printf("%lld\n",ans);
    return 0;
}

View Code