HDU 4825 Xor Sum(xor trie)

题意:

$N\le 10^5个数,M\le 10^5次询问$
$每次询问给出一个整数K,从数列A_i中找出一个数S使得K\oplus S值最大$

分析:

$xor trie基础题$
$将所有数插入01 trie中,每次询问就在trie上贪心的找$
$由于要最大,那么就从高位到低位插入,这样查找的时候也从高位到低位$
$由于xor的性质,0\oplus1=1,显然优先找不同的,这样贪心的找就可以了$
$时间复杂度为O(nb),b=log_2(max\{A_i\})$

代码:

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//
// Created by TaoSama on 2016-04-21
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, q;
struct Trie {
static const int M = 32 * 1e5 + 10, S = 2;
int root, sz;
int nxt[M][S], val[M];
int newNode() {
val[sz] = 0;
memset(nxt[sz], -1, sizeof nxt[sz]);
return sz++;
}
void init() {
sz = 0;
root = newNode();
}
void insert(int x) {
int u = root;
for(int i = 31; ~i; --i) {
int c = x >> i & 1, &v = nxt[u][c];
if(v == -1) v = newNode();
u = v;
}
val[u] = x;
}
int query(int x) {
int u = root;
for(int i = 31; ~i; --i) {
int c = x >> i & 1;
if(~nxt[u][c ^ 1]) u = nxt[u][c ^ 1];
else u = nxt[u][c];
}
return val[u];
}
} trie;
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &q);
trie.init();
for(int i = 1; i <= n; ++i) {
int x; scanf("%d", &x);
trie.insert(x);
}
static int kase = 0;
printf("Case #%d:\n", ++kase);
while(q--) {
int x; scanf("%d", &x);
printf("%d\n", trie.query(x));
}
}
return 0;
}


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