Description

P4146 序列终结者

Solution

fhq-treap

不得不说 \(fhq-treap\) 代码是真的短，真的好写。

挺板子的一道题。

• 区间加：打个 \(add\) 标记，不停下放即可，相应的 \(maxs\)，\(val\)，\(add\) 都要更新。

• 区间翻转：这就是个文艺平衡树板子，不多说了。（不会的话，建议先做一下板子）

• 查询区间最大值：把区间对应的子树找出来，输出这个子树记录的最大值即可。

注意 \(pushup\) 时要判断是否有左右子树，在向上传。

Code

#include <bits/stdc++.h>
#define ls(x) t[x].ch[0]
#define rs(x) t[x].ch[1]

using namespace std;

const int N = 5e4 + 10;
int n, m, root, tot;
struct Treap{
	int ch[2], siz, maxs, val, wei, add, lazy;
}t[N];
int a, b, c;

inline int read(){
	int x = 0, f = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
	return x * f;
}

inline void pushup(int x){
	t[x].siz = t[ls(x)].siz + t[rs(x)].siz + 1;
	t[x].maxs = t[x].val;
	if(ls(x)) t[x].maxs = max(t[x].maxs, t[ls(x)].maxs);
	if(rs(x)) t[x].maxs = max(t[x].maxs, t[rs(x)].maxs);
}

inline void pushdown(int x){
	if(!x) return;
	if(t[x].lazy){
		if(ls(x)) t[ls(x)].lazy ^= 1;
		if(rs(x)) t[rs(x)].lazy ^= 1;
		swap(ls(x), rs(x));
		t[x].lazy = 0;
	}
	if(t[x].add){
		if(ls(x)) t[ls(x)].val += t[x].add, t[ls(x)].add += t[x].add, t[ls(x)].maxs += t[x].add;
		if(rs(x)) t[rs(x)].val += t[x].add, t[rs(x)].add += t[x].add, t[rs(x)].maxs += t[x].add;
		t[x].add = 0;
	}
}

inline void split(int x, int k, int &a, int &b){
	if(!x){
		a = b = 0;
		return;
	}
	pushdown(x);
	if(k >= t[ls(x)].siz + 1){
		a = x;
		split(rs(x), k - t[ls(x)].siz - 1, rs(x), b);
	}else{
		b = x;
		split(ls(x), k, a, ls(x));
	}
	pushup(x);
}

inline int merge(int x, int y){
	if(!x || !y) return x | y;
	if(t[x].wei <= t[y].wei){
		pushdown(x);
		rs(x) = merge(rs(x), y);
		pushup(x);
		return x;
	}else{
		pushdown(y);
		ls(y) = merge(x, ls(y));
		pushup(y);
		return y;
	}
}

inline int newnode(int k){
	t[++tot].val = t[tot].maxs = k, t[tot].siz = 1, t[tot].wei = rand();
	return tot;
}

inline void insert(int k){
	root = merge(root, newnode(k));
}

inline void Add(int l, int r, int k){
	int len = r - l + 1;
	split(root, l - 1, a, b);
	split(b, len, b, c);
	t[b].maxs += k;
	t[b].val += k;
	t[b].add += k;
	root = merge(merge(a, b), c);
}

inline void Reverse(int l, int r){
	split(root, l - 1, a, b);
	split(b, r - l + 1, b, c);
	t[b].lazy ^= 1;
	root = merge(merge(a, b), c);
}

inline int Query(int l, int r){
	split(root, l - 1, a, b);
	split(b, r - l + 1, b, c);
	int ans = t[b].maxs;
	root = merge(merge(a, b), c);
	return ans;
}

int main(){
	n = read(), m = read();
	for(int i = 1; i <= n; i++)
		insert(0);
	while(m--){
		int op = read(), l = read(), r = read(), k;
		if(op == 1) k = read(), Add(l, r, k);
		else if(op == 2) Reverse(l, r);
		else printf("%d\n", Query(l, r));
	}
	return 0;
}

End