maximum clique 1
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld
题目描述
You are given a set S containing n distinct positive integers
a1,a2,…,an.
Please find a subset of S which has the maximum size while satisfying the following constraint:
The binary representations of any two numbers in this subset must have at least two different bits.
If there are multiple valid subsets, please output any one of them.
Please find a subset of S which has the maximum size while satisfying the following constraint:
The binary representations of any two numbers in this subset must have at least two different bits.
If there are multiple valid subsets, please output any one of them.
输入描述:
The input contains only two lines.
The first line contains an integer N denoting the size of S.
The second line contains N positive integers a1,a2,…,aN denoting the members of set S.
* 1≤N≤5000
* 1≤ai≤1e9
* all ai are distinct
输出描述:
You should output two lines.
The first line contains one integer m denoting the size of the subset you found.
The second line contains m positive integers denoting the member of this subset.
输入
5 3 1 4 2 5
输出
3 4 1 2
说明
3 (11
2) and 1 (01
2) has only 1 different bit so they can not be in the set together. 4 (100
2) and 1 (001
2) has 2 different bits so they can be in the set together.
Following unordered pairs are allowed to be in the set together: <1, 2>, <1, 4>, <2, 4>, <2, 5>, <3, 4>, <3, 5>. Thus the maximum set is of size 3 ({1, 2, 4}).
题意：在一个集合中求一个最大的子集，使得子集里任意两个数之间至少有两个二进制位不同。
思路：两个相异的数至少两个bit不一样的否命题是：恰一个bit不一样。然后考虑到A和B一个bit不一样，A和C一个bit不一样，A，B，C各不相同，那么B和C一定不会只有一个bit不一样，所以若是两个数只有一个bit不一样，我们就在它们之间连一条边，可以发现这个图是个二分图。答案就是这个二分图的最大点独立集。
#include<bits/stdc++.h> #define N 5050 using namespace std; typedef struct { int to,next; long long flow; bool is_match; }ss; ss edg[64*N]; int now_edge=0,s,t; int head[N]; void addedge(int u,int v) { edg[now_edge]=(ss){v,head[u],0}; head[u]=now_edge++; } void addedge(int u,int v,long long flow) { // printf("%d %d\n",u,v); edg[now_edge]=(ss){v,head[u],flow}; head[u]=now_edge++; edg[now_edge]=(ss){u,head[v],0}; head[v]=now_edge++; } int dis[N]; bool bfs() { memset(dis,0,sizeof(dis)); queue<int>q; q.push(s); dis[s]=1; while(!q.empty()) { int now=q.front(); q.pop(); for(int i=head[now];i!=-1;i=edg[i].next) { ss &e=edg[i]; if(e.flow>0&&dis[e.to]==0) { dis[e.to]=dis[now]+1; q.push(e.to); } } } if(dis[t]==0)return 0; return 1; } int current[N]; long long dfs(int x,long long maxflow) { if(x==t)return maxflow; for(int i=current[x];i!=-1;i=edg[i].next) { current[x]=i; ss &e=edg[i]; if(e.flow>0&&dis[e.to]==dis[x]+1) { long long flow=dfs(e.to,min(maxflow,e.flow)); if(flow!=0) { e.flow-=flow; edg[i^1].flow+=flow; return flow; } } } return 0; } long long dinic()//最大流 { long long ans=0,flow; while(bfs()) { for(int i=0;i<N;i++)current[i]=head[i]; while(flow=dfs(s,LLONG_MAX/2))ans+=flow; } return ans; } void init() { for(int i=0;i<N;i++)head[i]=-1; now_edge=0; } int arr[N]; int color[N]={0}; void Bipartite_graph_coloring(int x,int now_color)//二分图染色 { if(color[x])return; color[x]=now_color; for(int i=head[x];i!=-1;i=edg[i].next)Bipartite_graph_coloring(edg[i].to,now_color*-1); } bool is_match_vertex[N]={0}; int is_ans[N]={0}; void dfs(int x,int type)//寻找最小点覆盖集中的点，不在该集合中的点就是最大点独立集的点 { if(is_ans[x])return; is_ans[x]=1; for(int i=head[x];i!=-1;i=edg[i].next) { int v=edg[i].to; if(v==s||v==t)continue; if(edg[i].is_match==(bool)type)dfs(v,1-type); } } int main() { int n; scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d",&arr[i]); init(); for(int i=1;i<=n;i++) { for(int j=i+1;j<=n;j++) { int now=arr[i]^arr[j]; if(now==(now&-now)) { addedge(i,j); addedge(j,i); // printf("%d %d\n",arr[i],arr[j]); } } } for(int i=1;i<=n;i++) if(!color[i])Bipartite_graph_coloring(i,1);//先将图划分成二分图 /* for(int i=1;i<=n;i++)if(color[i]==1)printf("%d ",arr[i]); printf("\n"); for(int i=1;i<=n;i++)if(color[i]==-1)printf("%d ",arr[i]); printf("\n");*/ init(); s=n+1,t=s+1; for(int i=1;i<=n;i++) { if(color[i]==1)addedge(s,i,1); else addedge(i,t,1); if(color[i]==1) { for(int j=1;j<=n;j++) { int now=arr[i]^arr[j]; if(now==(now&-now)) { addedge(i,j,1);//建边准备跑二分图匹配 } } } } int ans=n-dinic(); printf("%d\n",ans); for(int i=1;i<=n;i++) { if(color[i]==1) { for(int j=head[i];j!=-1;j=edg[j].next) { if(edg[j].to!=s&&edg[j^1].flow) { edg[j].is_match=edg[j^1].is_match=true; is_match_vertex[i]=is_match_vertex[edg[j].to]=true; } else { edg[j].is_match=edg[j^1].is_match=false; } } } } for(int i=1;i<=n;i++) { if(color[i]==-1&&is_match_vertex[i]==false) { dfs(i,0); } } for(int i=1;i<=n;i++) { if(color[i]==1&&!is_ans[i]||color[i]==-1&&is_ans[i])printf("%d%c",arr[i],--ans==0 ? '\n': ' '); } return 0; }