HDU 5536 Chip Factory(xor trie)

题意:

$3\le N\le 10^3个数,求\max_{i,j,k} (s_i+s_j) \oplus s_k的最大值,i\ne j\ne k$

分析:

$xor trie简单技巧$
$将所有数插入01 trie中,n^2枚举s_i+s_j,由于要不同先把它俩删了$
$然后再去贪心的找s_k,不要忘记恢复就好$
$时间复杂度O(n^2b),b=log_2(max\{A_i\})$

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, a[N];
struct Trie {
static const int M = 32 * 1e3 + 10, S = 2;
int root, sz;
int nxt[M][S], have[M];
int newNode() {
have[sz] = 0;
memset(nxt[sz], -1, sizeof nxt[sz]);
return sz++;
}
void init() {
sz = 0;
root = newNode();
}
void update(int x, int d) {
int u = root;
for(int i = 31; ~i; --i) {
int c = x >> i & 1, &v = nxt[u][c];
if(v == -1) v = newNode();
have[v] += d;
u = v;
}
}
int query(int x) {
int u = root, ret = 0;
for(int i = 31; ~i; --i) {
int c = x >> i & 1;
if(have[nxt[u][c ^ 1]]) {
ret |= 1 << i;
u = nxt[u][c ^ 1];
} else u = nxt[u][c];
}
return ret;
}
} 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", &n);
trie.init();
for(int i = 1; i <= n; ++i) {
scanf("%d", a + i);
trie.update(a[i], 1);
}
int ans = 0;
for(int i = 1; i <= n; ++i) {
trie.update(a[i], -1);
for(int j = i + 1; j <= n; ++j) {
trie.update(a[j], -1);
ans = max(ans, trie.query(a[i] + a[j]));
trie.update(a[j], 1);
}
trie.update(a[i], 1);
}
printf("%d\n", ans);
}
return 0;
}


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