CodeForces - 165E Compatible Numbers(SOSdp)

题目链接：点击查看

题目大意：给出 nnn 个数，问能否在数列中找到一个数，满足 ai&aj=0a_i\& a_j=0ai&aj=0

题目分析：题目中的式子可以转换为，aia_iai 是 aja_jaj 补集的子集，然后就是 SOSdpSOSdpSOSdp 的模板题了，因为输出的答案涉及到了 −1-1−1，所以可以将 dpdpdp 初始化为 −1-1−1，然后直接维护每个超集的最大可行答案就好啦

转移的话就是子集向超集转移

代码：

// Problem: E. Compatible Numbers
// Contest: Codeforces - Codeforces Round #112 (Div. 2)
// URL: https://codeforces.com/contest/165/problem/E
// Memory Limit: 256 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) x&-x
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{T f=1;x=0;char ch=getchar();while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();x*=f;
}
template<typename T>
inline void write(T x)
{if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1<<22;
int a[N],dp[N];
int main()
{#ifndef ONLINE_JUDGE
//  freopen("data.in.txt","r",stdin);
//  freopen("data.out.txt","w",stdout);
#endif
//  ios::sync_with_stdio(false);memset(dp,-1,sizeof(dp));int n;read(n);for(int i=1;i<=n;i++) {read(a[i]);dp[a[i]]=a[i];}for(int j=0;j<22;j++) {for(int i=0;i<1<<22;i++) {//子集向超集转移if((i>>j)&1) {dp[i]=max(dp[i],dp[i^(1<<j)]);}}}for(int i=1;i<=n;i++) {printf("%d ",dp[(N-1)^a[i]]);}return 0;
}

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