ifrog-1028 Bob and Alice are playing numbers(trie树)

Posted 2020-08-07 LittlePointer

题目链接：

Bob and Alice are playing numbers

DESCRIPTION

Bob and his girl friend are playing game together.This game is like this: There are $\mathrm{nn\; numbers.\; If\; op\; =\; 11,Bob\; wants\; to\; find\; two\; numbers\; aiai\; and\; ajaj,that\; aiai\; \&\; ajaj\; will\; become\; maximum\; value.\; If\; op\; =\; 22,Bob\; wants\; to\; find\; two\; numbers\; aiai\; and\; ajaj,that\; aiai\; ^\; ajaj\; will\; become\; maximum\; value.\; If\; op\; =\; 33,Bob\; wants\; to\; find\; two\; numbers\; aiai\; and\; ajaj,that\; aiai\; |\; ajaj\; will\; become\; maximum\; value.\; Notice:\; \&\; for\; bitwise\; AND.\; (4\; \&\; 2)\; is\; 0,\; (4\; \&\; 5)\; is\; 4.\; ^\; for\; bitwise\; XOR.\; (4\; ^\; 2)\; is\; 6,\; (4\; ^\; 5)\; is\; 1.\; |\; for\; bitwise\; OR\; .\; (4\; |\; 2)\; is\; 6,\; (4\; |\; 5)\; is\; 5.}$

INPUT

First line is a positive integer $\mathrm{TT\; ,\; represents\; there\; are\; TT\; test\; cases.\; For\; each\; test\; case:\; First\; line\; includes\; two\; numbers\; n(2\le n\le 100000),op(1\le op\le 3)n(2\le n\le 100000),op(1\le op\le 3).\; The\; next\; line\; contains\; nn\; numbers:\; a1,a2,...,an(1\le ai\le 1000000)a1,a2,...,an(1\le ai\le 1000000).}$

OUTPUT

For the $\mathrm{ii-th\; test\; case\; ,\; first\; output\; Case\; \#i:\; in\; a\; single\; line.\; Then\; output\; the\; answer\; of\; ii-th\; test\; case.}$

SAMPLE INPUT

 

3

2 1

4 2

2 2

4 2

2 3

4 2

SAMPLE OUTPUT

 

Case #1: 0

Case #2: 6

Case #3: 6

 

题意：

 

^ & |三种操作找到能得到的最大值;

 

思路：

 

异或的值就是经典的trie树,&可以转化成trie树上的贪心,|也是一个贪心了,对于一个数x[i]，那么就找它所有为0的位置,看最大能贪心多少;

 

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
  
using namespace std;
  
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
  
typedef  long long LL;
  
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
    for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + ‘0‘);
    putchar(‘\n‘);
}
  
const int mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=(1<<20)+10;
const int maxn=1e5+110;
const double eps=1e-12;
 

int ch[21*maxn][2],sz,val[21*maxn],s[23],dp[21],x[maxn],n;
bool vis[N*2];
void insert(int x)
{
    mst(s,0);
    int n=0,u=0;
    while(x)s[n++]=(x&1),x>>=1;
    for(int i=20;i>=0;i--)
    {
        int c=s[i];
        if(!ch[u][c])
        {
            ch[sz][0]=ch[sz][1]=0;
            ch[u][c]=sz++;
        }
        u=ch[u][c];
        val[u]++;
    }
}
inline int query_or()
{
    memset(vis,false,sizeof(vis));
    For(i,1,n)vis[x[i]]=true;
    for(int i=(1<<20);i>0;i--)
    {
        for(int j=0;j<=20;j++)
        {
            if(!(i&(1<<j)))vis[i]|=vis[i|(1<<j)];
        }
    }
    int ans=0,t[22];
    for(int i=1;i<=n;i++)
    {
        int num=0,temp=0;
        for(int j=20;j>=0;j--)
        {
            if(!(x[i]&(1<<j)))t[num++]=(1<<j);
        }
        for(int j=0;j<num;j++)
        {
            if(vis[temp|t[j]])temp|=t[j];
        }
        ans=max(ans,temp|x[i]);
    }
    return ans;
}
void same(int &l,int r)
{
    if(!r)return ;
    if(!l)
    {
        ch[sz][0]=ch[sz][1]=0;
        l=sz++;
    }
    val[l]+=val[r];
    same(ch[l][0],ch[r][0]);
    same(ch[l][1],ch[r][1]);
}
inline int query_and()
{
    int u=0,ans=0;
    for(int i=20;i>=0;i--)
    {
        if(val[ch[u][1]]>=2)u=ch[u][1],ans|=dp[i];
        else same(ch[u][0],ch[u][1]),u=ch[u][0];
    }
    return ans;
}
inline int query_xor(int x)
{
    mst(s,0);
    int n=0,u=0,ans=0;
    while(x)s[n++]=(x&1),x>>=1;
    for(int i=20;i>=0;i--)
    {
        int c=s[i];
        if(ch[u][c^1])ans|=dp[i],u=ch[u][c^1];
        else u=ch[u][c];
    }
    return ans;
}
inline void Init()
{
    sz=1;
    mst(ch[0],0);
    mst(val,0);
    for(int i=0;i<=20;i++)
        dp[i]=(1<<i);
}
int main()
{
      int t,Case=0;
      read(t);
      while(t--)
      {
            int op,ans=0;
            read(n);read(op);
            Init();
            read(x[1]);insert(x[1]);
            For(i,2,n)
            {
                read(x[i]);
                if(op==2)ans=max(ans,query_xor(x[i]));
                if(op<3)insert(x[i]);
            }
            if(op==1)ans=query_and();
            else if(op==3)ans=query_or();
            printf("Case #%d: %d\n",++Case,ans);
      }
    return 0;
}