independent set 1

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 102400K，其他语言204800K
64bit IO Format: %lld

题目描述

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.

输入描述:

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.

输出描述:

Print one line, containing one integer represents the answer.

输入

输出

说明

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.
链接：https://ac.nowcoder.com/acm/contest/885/E
来源：牛客网

题意：求一个图的2^n种子图的最大点独立集。
思路：

•我们可以用一个 n-bit 2 进制整数来表示一个点集，第 i 个 bit 是 1 就代表点集包含第 i 个点，若是 0 则不包含
• 每个点相邻的点也可以用一个 n-bit 2 进制整数表示，计做 ci，若第 i 个点和第 j 个点相邻， ci 的第 j 个 bit 是 1，否则是 0
• 记 x 的最低位且是 1 的 bit 的位置是 lbx
• 令 dp[x] 代表点集 x 的最大独立集 size，那么我们能够根据点 lbx 是否属于最大独立集来列出以下关系式：
dp[x] = max(dp[x - (1<<lbx)], dp[x & (~clb_x)] + 1) (使用 c 语言运算符)

•高效位运算参考博客：https://blog.csdn.net/yuer158462008/article/details/46383635

#include<bits/stdc++.h>

using namespace std;

char dp[<<];

int Map[]={};

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]|=(<<v);

        Map[v]|=(<<u);

    }

    for(int i=;i<n;i++)Map[i]|=<<i;

    int upper=<<n;

    long long ans=;

    for(int i=;i<upper;i++)

    {

        int lbx=__builtin_ctz(i);

        dp[i] = max(dp[i - (<<lbx)] , dp[i & (~Map[lbx])] + );

        ans+=dp[i];

    }

    printf("%lld\n",ans);

    return ;

}