HDU5536 Chip Factory

题意:

给出一个数组\(s\),求

\[max_{i,j,k}(s_i + s_j)\oplus s_k ,i\neq j\neq k \]

思路:

01字典树,首先还是正常插入。可以想到枚举\(i\)\(j\)的和,再字典树跑\(k\),这里涉及下标不能相同,所以可以把\(i\)\(j\),先删除了。在插入的时候计数\(cnt[u]++\),删除就依次把到达的各点\(cnt[u]--\)。跑\(k\)的时候判断是否可到达即可。

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = 0X3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1000000007;
typedef long long ll;

int _;
int n;
int a[1010];
int tri[32 * 1010][2];
int cnt[32 * 1010], val[32 * 1010];
int tot;

void init() {
    memset(tri, 0, sizeof(tri));
    memset(val, 0, sizeof(val));
    memset(cnt, 0, sizeof(cnt));
    tot = 1;
}

void Insert(int x) {
    int u = 0;
    for (int i = 31; i >= 0; i--) {
        int bit = (x & (1 << i)) ? 1 : 0;
        if (!tri[u][bit])
            tri[u][bit] = tot++;
        u = tri[u][bit];
        cnt[u]++;
    }
    val[u] = x;
}

void Del(int x) {
    int u = 0;
    for (int i = 31; i >= 0; i--) {
        int bit = (x & (1 << i)) ? 1 : 0;
        u = tri[u][bit];
        cnt[u]--;
    }
}

int Query(int x) {
    int u = 0;
    for (int i = 31; i >= 0; i--) {
        int bit = (x & (1 << i)) ? 1 : 0;
        if (tri[u][bit ^ 1] && cnt[tri[u][bit ^ 1]] > 0) {
            u = tri[u][bit ^ 1];
        } else {
            u = tri[u][bit];
        }
    }
    return x ^ val[u];
}

int main() {
    for (scanf("%d", &_); _; _--) {
        init();
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            Insert(a[i]);
        }
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            Del(a[i]);
            for (int j = i + 1; j <= n; j++) {
                Del(a[j]);
                ans = max(ans, Query(a[i] + a[j]));
                Insert(a[j]);
            }
            Insert(a[i]);
        }
        printf("%d\n", ans);
    }
}

posted @ 2020-01-21 18:18  Frontierone  阅读(105)  评论(0编辑  收藏  举报