【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 longusing 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;
}

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【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

思路

代码

【luogu P5055】【模板】可持久化文艺平衡树相关推荐

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