【CodeChef-XRQRS】Xor Queries【可持久化Trie / +主席树】

【题目链接】

有中文题面就不发题意了。

似乎维护一个可持久化Trie和一个主席树就可以做了,但是仔细想想好像只需要一个可持久化Trie就完了。

脑补了一下Trie上找第k大和统计数个数,似乎是对了。

1A了。。

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

using namespace std;

const int maxn = 500005, 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();
		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/51217012
    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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