[HG]腿部挂件 题解

前言

暴力跑的比正解快。
以下暴力(循环展开+fread读入输出优化)

#include<cstdio>
#pragma GCC optimize(3, "Ofast")

int a[200010];

namespace fast_IO{
    const int IN_LEN = 10000000, OUT_LEN = 10000000;
    char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
    inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
    inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
    inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
    inline int read(){
        int x = 0; char ch = ' ';
        while (ch < '0' || ch > '9') ch = getchar_();
        while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_(); return x;
    }
    inline void write(int x){
        if (x < 0) putchar_('-'), x = -x;
        if (x > 9) write(x / 10);
        putchar_(x % 10 + 48);
    }
}

using namespace fast_IO;

#define max(a,b) (a>(b)?a:(b))

int main(){
    const int n = read(); int q = read();
    for (register int i = 1; i <= n; ++i)
        a[i] = read();
    while (q--){
        const int x = read(), l = read(), rr = read();
        register int ans = 0;
        const int lft = ((rr - l) >> 2 << 2) + l;
        for (register int j = l; j <= lft - 4; j += 4){
            (ans < (x ^ a[j + 1])) && (ans = x ^ a[j + 1]);
            (ans < (x ^ a[j + 2])) && (ans = x ^ a[j + 2]);
            (ans < (x ^ a[j + 3])) && (ans = x ^ a[j + 3]);
            (ans < (x ^ a[j + 4])) && (ans = x ^ a[j + 4]);
        }
        for (register int j = lft; j <= rr; ++j)
            ans = max(ans, x ^ a[j + 1]);
        write(ans), putchar_('\n');
    }
    flush(); return 0;
}

题目

给定一个长度为 \(n(1 \leq n \leq 2 \times 10^5)\) 的数列，
给定 \(q(1 \leq q \leq 2 \times 10^5)\) 个询问，
每个询问给定三个整数 \(x, l, r\) 。
在区间 \([l,r]\) 中选定一个数，使它与 \(x\) 的异或值最大。
查询

题解

如果是查询区间是一个固定区间，很容易想到，使用trie树。
那么如果查询不是固定区间，那么显然珂以使用可持久化trie树切掉此题，
预计时间复杂度\(\Theta(n\ logn)\)
但是博主才学疏浅，不会这个算法。
就这么结束了吗？
并没有，显然我们可以用类线段树结构维护trie树，复杂度\(\Theta(n\ log^2n)\)
考场上常数过大 一不小心 TLE了
后来一边query一边建树，就AC了。

代码

#pragma GCC optimize(3, "Ofast")
#include <cstdio>

namespace fast_IO{
    const int IN_LEN = 10000000, OUT_LEN = 10000000;
    char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
    inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
    inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
    inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
    int read(){
        int x = 0; char ch = ' ';
        while (ch < '0' || ch > '9') ch = getchar_();
        while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_(); return x;
    }
    void write(int x){
        if (x < 0) putchar_('-'), x = -x;
        if (x > 9) write(x / 10);
        putchar_(x % 10 + '0');
    }
}

using namespace fast_IO;

struct Trie{
    int c[2];
} trie[57108864];
int cnt = 0;

void insert(int x, int rt){
    for (register int i = 1 << 29; i; i >>= 1){
        bool to = x & i;
        if (trie[rt].c[to] == -1){
            trie[rt].c[to] = ++cnt;
            trie[cnt].c[0] = trie[cnt].c[1] = -1;
        }
        rt = trie[rt].c[to];
    }
}

int query(int x, int rt){
    int res = 0;
    for (register int i = 1 << 29; i; i >>= 1){
        bool to = (x ^ i) & i;
        if (~trie[rt].c[to]){
            res += i;
            rt = trie[rt].c[to];
        }
        else rt = trie[rt].c[to ^ 1];
    }
    return res;
}

int a[200005];

struct Seg{
    int root;
    int lc, rc;
} seg[800005];

inline int max(int a, int b){
    return ((a > b) ? a : b);
}

int querySeg(int pos, int l, int r, int x, int y, int v){
    if (x <= l && r <= y){
        if (!seg[pos].root){
            seg[pos].root = ++cnt;
            trie[cnt].c[0] = trie[cnt].c[1] = -1;
            for (register int i = l; i <= r; ++i)
                insert(a[i], seg[pos].root);
        }
        return query(v, seg[pos].root);
    }
    int mid = l + r >> 1, res = 0;
    if (x <= mid)
        res = querySeg(pos << 1, l, mid, x, y, v);
    if (y > mid)
        res = max(res, querySeg(pos << 1 | 1, mid + 1, r, x, y, v));
    return res;
}

int main(){
    int n = read(), q = read();
    trie[0].c[0] = trie[0].c[1] = -1;
    for (register int i = 1; i <= n; ++i)
        a[i] = read();
    while (q--){
        int x = read(), l = read() + 1, r = read() + 1;
        write(querySeg(1, 1, n, l, r, x)), putchar_('\n');
    }
    flush(); return 0;
}

标签：ch,oh,int,题解,obuf,LEN,trie,挂件,HG
来源： https://www.cnblogs.com/linzhengmin/p/11764932.html