首页 > 代码库 > ifrog-1028 Bob and Alice are playing numbers(trie树)
ifrog-1028 Bob and Alice are playing numbers(trie树)
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;}
ifrog-1028 Bob and Alice are playing numbers(trie树)
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。