【BZOJ4546】codechef XRQRS【可持久化Trie / +主席树】

【题目链接】

早上刚在CC上做了,中午就被人搬到BZOJ了。。

不知道为什么n开50wRE,开了52w卡了下内存就过了。

【CC上这题题解】

/* Pigonometry */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 520005, maxk = 21, maxnode = maxn * maxk;

int root[maxn], triecnt, son[maxnode][2], sum[maxnode], bin[maxk];

inline int iread() {
    int f = 1, x = 0; char ch = getchar();
    for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
    for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
    return f * x;
}

inline void insert(int &pos, int c) {
    int now = ++triecnt, last = pos; pos = now;
    for(int i = maxk - 1; i >= 0; i--) {
        son[now][0] = son[last][0]; son[now][1] = son[last][1];
        sum[now] = sum[last] + 1;
        bool ind = c & bin[i];
        last = son[last][ind];
        now = son[now][ind] = ++triecnt;
    }
    son[now][0] = son[last][0]; son[now][1] = son[last][1];
    sum[now] = sum[last] + 1;
}

inline int query(int last, int now, int c) {
    int ans = 0;
    for(int i = maxk - 1; i >= 0; i--) {
        bool ind = c & bin[i];
        if(sum[son[now][ind ^ 1]] - sum[son[last][ind ^ 1]] > 0)
            now = son[now][ind ^ 1], last = son[last][ind ^ 1], ans |= (ind ? 0 : bin[i]);
        else
            now = son[now][ind], last = son[last][ind], ans |= (ind ? bin[i] : 0);
    }
    return ans;
}

inline int findless(int last, int now, int c) {
    int ans = 0;
    for(int i = maxk - 1; i >= 0; i--) {
        bool ind = c & bin[i];
        if(ind) ans += sum[son[now][0]] - sum[son[last][0]];
        if(sum[son[now][ind]] - sum[son[last][ind]] == 0) return ans;
        now = son[now][ind]; last = son[last][ind];
    }
    ans += sum[now] - sum[last];
    return ans;
}

inline int findkth(int last, int now, int k) {
    int ans = 0;
    for(int i = maxk - 1; i >= 0; i--) {
        int tmp = sum[son[now][0]] - sum[son[last][0]];
        if(tmp >= k) now = son[now][0], last = son[last][0];
        else k -= tmp, now = son[now][1], last = son[last][1], ans |= bin[i];
    }
    return ans;
}

int main() {
    bin[0] = 1;
    for(int i = 1; i < maxk; i++) bin[i] = bin[i - 1] << 1;

    int n = 0;
    for(int T = iread(); T; T--) {
        int opt = iread(); opt--;
        if(opt == 0) {
            int x = iread(); n++;
            insert(root[n] = root[n - 1], x);
        }
        else if(opt == 1) {
            int l = iread(), r = iread(), x = iread();
            printf("%d\n", query(root[l - 1], root[r], x));
        }
        else if(opt == 2) {
            int k = iread(); n -= k;
            triecnt = root[n + 1] - 1;
        }
        else if(opt == 3) {
            int l = iread(), r = iread(), x = iread();
            printf("%d\n", findless(root[l - 1], root[r], x));
        }
        else if(opt == 4) {
            int l = iread(), r = iread(), k = iread();
            printf("%d\n", findkth(root[l - 1], root[r], k));
        }
    }
    return 0;
}
    原文作者:Trie树
    原文地址: https://blog.csdn.net/braketbn/article/details/51219856
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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