【luogu P5055】【模板】可持久化文艺平衡树

【模板】可持久化文艺平衡树

题目链接：luogu P5055

题目大意

要你维护插入，删除，区间翻转，区间求和。
但要求可持续化，即每次操作在一个历史版本上进行，且会产生一个新的历史版本

思路

看到题目都是可持续化平衡树了。

直接上 splay，这题要多搞一个 down。
down 交换左右儿子之前要复制左右儿子。

然后其他的就跟 可持久化平衡树 是差不多的了。

代码

#include<queue>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define ll long long

using namespace std;

int n, rt[200001], bb, op, x, y, tot;
int ls[200001 << 7], rs[200001 << 7], yj[200001 << 7], sz[200001 << 7];
bool turn[200001 << 7];
ll sum[200001 << 7], lastans, val[200001 << 7];

int newpoint(ll num) {
	int re = ++tot;
	ls[re] = rs[re] = turn[re] = 0;
	val[re] = num;
	sum[re] = num;
	yj[re] = rand();
	sz[re] = 1;
	return re;
}

int copypoint(int pl) {
	int re = ++tot;
	ls[re] = ls[pl];
	rs[re] = rs[pl];
	val[re] = val[pl];
	sum[re] = sum[pl];
	yj[re] = yj[pl];
	turn[re] = turn[pl];
	sz[re] = sz[pl];
	return re;
}

void up(int now) {
	sz[now] = sz[ls[now]] + sz[rs[now]] + 1;
	sum[now] = sum[ls[now]] + sum[rs[now]] + val[now];
}

void down(int now) {
	if (turn[now]) {
		if (ls[now]) ls[now] = copypoint(ls[now]);//记得翻转前左右儿子要复制
		if (rs[now]) rs[now] = copypoint(rs[now]);
		swap(ls[now], rs[now]);
		if (ls[now]) turn[ls[now]] ^= 1;
		if (rs[now]) turn[rs[now]] ^= 1;
		turn[now] ^= 1;
	}
} 

pair <int, int> split_rnk(int now, int rnk) {
	if (!now) return make_pair(0, 0);
	if (!rnk) return make_pair(0, copypoint(now));
	
	down(now);
	
	pair <int, int> re;
	if (rnk <= sz[ls[now]]) {
		int noww = copypoint(now);
		re = split_rnk(ls[noww], rnk);
		ls[noww] = re.second;
		up(noww);
		re.second = noww;
	}
	else {
		int noww = copypoint(now);
		re = split_rnk(rs[noww], rnk - sz[ls[now]] - 1);
		rs[noww] = re.first;
		up(noww);
		re.first = noww;
	}
	
	return re;
}

int merge(int x, int y) {
	if (x) down(x);
	if (y) down(y);
	if (!x) return y;
	if (!y) return x;
	
	if (yj[x] < yj[y]) {
		int xx = copypoint(x);
		rs[xx] = merge(rs[xx], y);
		up(xx);
		return xx;
	}
	else {
		int yy = copypoint(y);
		ls[yy] = merge(x, ls[yy]);
		up(yy);
		return yy;
	}
}

void insert(int bb, int pl, int num) {
	pair <int, int> x = split_rnk(rt[bb], pl);
	int y = newpoint(num);
	rt[bb] = merge(x.first, merge(y, x.second));
}

void delete_(int bb, int pl) {
	pair <int, int> x = split_rnk(rt[bb], pl - 1);
	pair <int, int> y = split_rnk(x.second, 1);
	rt[bb] = merge(x.first, y.second);
}

void turn_(int bb, int l, int r) {
	pair <int, int> x = split_rnk(rt[bb], l - 1);
	pair <int, int> y = split_rnk(x.second, r - l + 1);
	
	turn[y.first] ^= 1;
	
	rt[bb] = merge(merge(x.first, y.first), y.second);
}

ll get_sum(int bb, int l, int r) {
	pair <int, int> x = split_rnk(rt[bb], l - 1);
	pair <int, int> y = split_rnk(x.second, r - l + 1);
	return sum[y.first];
}

int main() {
	srand(19491001);
	
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d %d", &bb, &op);
		rt[i] = rt[bb];
		if (op == 1) {
			scanf("%d %d", &x, &y);
			x ^= lastans; y ^= lastans;
			insert(i, x, y); 
			continue;
		}
		if (op == 2) {
			scanf("%d", &x);
			x ^= lastans;
			delete_(i, x);
			continue;
		}
		if (op == 3) {
			scanf("%d %d", &x, &y);
			x ^= lastans;
			y ^= lastans;
			turn_(i, x, y);
			continue;
		}
		if (op == 4) {
			scanf("%d %d", &x, &y);
			x ^= lastans;
			y ^= lastans;
			lastans = get_sum(i, x, y);
			printf("%lld\n", lastans);
		}
	}
	
	return 0;
}

__EOF__

本文作者：あおいSakura
本文链接：https://www.cnblogs.com/Sakura-TJH/p/luogu_P5055.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！