Codeforces 282E Sausage Maximization(字典树)

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题目链接：282E Sausage Maximization

题目大意：给定一个序列A，要求从中选取一个前缀，一个后缀，可以为空，当时不能重叠，亦或和最大。

解题思路：预处理出前缀后缀亦或和，然后在字典树中维护，每次添加并查询，过程中维护ans。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

typedef long long ll;
const int maxn = 1e5 + 5;

struct Tire {
    int sz, g[maxn * 100][2];

    void init();
    void insert(ll s);
    ll find(ll s);
}T;

int N;
ll A[maxn], prefix[maxn], suffix[maxn];

int main () {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++)
        scanf("%lld", &A[i]);

    for (int i = 1; i <= N; i++)
        prefix[i] = prefix[i-1] ^ A[i];
    for (int i = N; i; i--)
        suffix[i] = suffix[i+1] ^ A[i];

    ll ans = 0;
    T.init();

    for (int i = N; i >= 0; i--) {
        T.insert(suffix[i+1]);
        ans = max(ans, T.find(prefix[i]));
    }
    printf("%lld\n", ans);
    return 0;
}

void Tire::init() {
    sz = 1;
    memset(g[0], 0, sizeof(g[0]));
}

void Tire::insert(ll s) {
    int u = 0;

    for (int i = 60; i >= 0; i--) {
        int v = (s>>i)&1;

        if (g[u][v] == 0) {
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }
}

ll Tire::find (ll s) {
    int u = 0;
    ll ret = 0;

    for (int i = 60; i >= 0; i--) {
        int v = ((s>>i)&1) ^ 1;

        if (g[u][v])
            ret |= (1LL<<i);
        else
            v = v ^ 1;
        u = g[u][v];
    }
    return ret;
}

Codeforces 282E Sausage Maximization(字典树)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/40630195