[luoguP3332] [ZJOI2013]K大数查询（树套树）

传送门

 

一开始想的是区间线段树套权值线段树，结果好像不能实现。

然后题解是权值线段树套区间线段树。

区间线段树上标记永久化就省去了pushdown的操作减少常数。

标记永久化的话。。yy不出来就看代码吧。

然后注意开long long

 

#include <cstdio>
#include <iostream>
#include <algorithm>
#define N 50010
#define LL long long

using namespace std;

int n, m, t, cnt;
LL sum[N * 200];
int opt[N], a[N], b[N], c[N], g[N], add[N * 200], ls[N * 200], rs[N * 200], root[N << 2];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void insert2(int &now, int l, int r, int x, int y)
{
	if(!now) now = ++cnt;
	if(x <= l && r <= y)
	{
		add[now]++;
		sum[now] += r - l + 1;
		return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) insert2(ls[now], l, mid, x, y);
	if(mid < y) insert2(rs[now], mid + 1, r, x, y);
	sum[now] = sum[ls[now]] + sum[rs[now]] + (LL)(r - l + 1) * add[now];
}

inline void insert1(int now, int l, int r, int d, int x, int y)
{
	insert2(root[now], 1, n, x, y);
	if(l == r) return;
	int mid = (l + r) >> 1;
	if(d <= mid) insert1(now << 1, l, mid, d, x, y);
	else insert1(now << 1 | 1, mid + 1, r, d, x, y);
}

inline LL query2(int now, int l, int r, int x, int y)
{
	if(x <= l && r <= y) return sum[now];
	LL tmp = 0;
	int mid = (l + r) >> 1;
	if(x <= mid) tmp += query2(ls[now], l, mid, x, y);
	if(mid < y) tmp += query2(rs[now], mid + 1, r, x, y);
	return tmp + (LL)(min(r, y) - max(l, x) + 1) * add[now];
}

inline int query1(int now, int l, int r, LL d, int x, int y)
{
	if(l == r) return l;
	int mid = (l + r) >> 1;
	LL tmp = query2(root[now << 1 | 1], 1, n, x, y);
	if(tmp < d) return query1(now << 1, l, mid, d - tmp, x, y);
	else return query1(now << 1 | 1, mid + 1, r, d, x, y);
}

int main()
{
	int i;
	n = read();
	m = read();
	for(i = 1; i <= m; i++)
	{
		opt[i] = read();
		a[i] = read(), b[i] = read(), c[i] = read();
		if(opt[i] == 1) g[++t] = c[i];
	}
	sort(g + 1, g + t + 1);
	t = unique(g + 1, g + t + 1) - g - 1;
	for(i = 1; i <= m; i++)
		if(opt[i] == 1)
		{
			c[i] = lower_bound(g + 1, g + t + 1, c[i]) - g;
			insert1(1, 1, t, c[i], a[i], b[i]);
		}
		else printf("%d\n", g[query1(1, 1, t, c[i], a[i], b[i])]);
	return 0;
}