2019 牛客多校五 F. maximum clique 1 (最大团)

大意: 给定$n$个互不相同的数, 若两个数异或后二进制中$1$的个数不少于$2$则连边, 求最大团.

 

最大团转为补图最大独立集. 可以发现补图是二分图, 所以直接$dinic$即可.

最大独立集相当于n-最小割, 最终$X$部仍与$S$相连的点和$Y$部不与$S$相连的点构成最大独立集.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
const int N = 1e6+10, S = N-2, T = N-1, INF = 0x3f3f3f3f;
int n, a[N], b[N];
struct edge {
    int to,w,next;
    edge(int to=0,int w=0,int next=0):to(to),w(w),next(next){}
} e[N];
int head[N], dep[N], vis[N], cur[N], cnt=1;
queue<int> Q;
int bfs() {
    REP(i,1,n) dep[i]=INF,vis[i]=0,cur[i]=head[i];
    dep[S]=INF,vis[S]=0,cur[S]=head[S];
    dep[T]=INF,vis[T]=0,cur[T]=head[T];
    dep[S]=0,Q.push(S);
    while (Q.size()) {
        int u = Q.front(); Q.pop();
        for (int i=head[u]; i; i=e[i].next) {
            if (dep[e[i].to]>dep[u]+1&&e[i].w) {
                dep[e[i].to]=dep[u]+1;
                Q.push(e[i].to);
            }
        }
    }
    return dep[T]!=INF;
}
int dfs(int x, int w) {
    if (x==T) return w;
    int used = 0;
    for (int i=cur[x]; i; i=e[i].next) {
        cur[x] = i;
        if (dep[e[i].to]==dep[x]+1&&e[i].w) {
            int f = dfs(e[i].to,min(w-used,e[i].w));
            if (f) used+=f,e[i].w-=f,e[i^1].w+=f;
            if (used==w) break;
        }
    }
    return used;
}
int dinic() {
    int ans = 0;
    while (bfs()) ans+=dfs(S,INF);
    return ans;
}
void add(int u, int v, int w) {
    e[++cnt] = edge(v,w,head[u]);
    head[u] = cnt;
    e[++cnt] = edge(u,0,head[v]);
    head[v] = cnt;
} 

int main() {
	scanf("%d", &n);
	REP(i,1,n) { 
		scanf("%d",a+i);
		b[i] = __builtin_parity(a[i]);
		if (b[i]) add(S,i,1);
		else add(i,T,1);
	}
	REP(i,1,n) REP(j,i+1,n) {
		int x = a[i]^a[j];
		if ((x&(x-1))==0) {
			if (b[i]) add(i,j,INF);
			else add(j,i,INF);
		}
	}
	printf("%d\n",n-dinic());
	REP(i,1,n) if ((dep[i]==INF)^b[i]) printf("%d ", a[i]);hr;
}