正文
ifrog-1028 Bob and Alice are playing numbers(trie树)
小程序：扫一扫查出行【扫一扫了解最新限行尾号】
复制小程序
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 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 TT , represents there are TT test cases. For each test case: First line includes two numbers n(2≤n≤100000),op(1≤op≤3)n(2≤n≤100000),op(1≤op≤3). The next line contains nn numbers: a1,a2,...,an(1≤ai≤1000000)a1,a2,...,an(1≤ai≤1000000).
OUTPUT
For the 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;
}
