【ybt金牌导航6-6-2】动态排名（整体二分）（树状数组）

动态排名

题目链接：ybt金牌导航6-6-2

题目大意

给你一个数组，然后会有两种操作：
修改一个位置的值和查询区间第 k 大。

思路

这题其实就跟不带修改的差不多。
（你修改你可以相当于一个删除再加入）

然后你就可以用整体二分做（数组数组维护排名）。

代码

#include<cstdio>

using namespace std;

struct node {
	int id, x, y, k, op, tmp;
}q[150001], q1[150001], q2[150001];
int n, m, x, num, y, k, tot;
int ans[10001], a[50001], tmp[150001];
char op;

struct SZSJ {
	int tr[150001];
	
	void insert(int x, int y) {
		for (; x <= n; x += x & (-x))
			tr[x] += y;
	}
	
	int query(int x) {
		int re = 0;
		for (; x; x -= x & (-x))
			re += tr[x];
		return re;
	}
}T;

void work(int l, int r, int L, int R) {
	if (l > r) return ;
	if (L == R) {
		for (int i = l; i <= r; i++)
			if (q[i].op == 1) ans[q[i].id] = L;
		return ;
	}
	int mid = (1ll * L + R) >> 1;
	for (int i = l; i <= r; i++) {
		if (q[i].op == 0 && q[i].y <= mid) T.insert(q[i].x, 1);
		if (q[i].op == 2 && q[i].y <= mid) T.insert(q[i].x, -1);
		if (q[i].op == 1) {
			tmp[i] = T.query(q[i].y) - T.query(q[i].x - 1);
		}
	}
	int l1 = 0, l2 = 0;
	for (int i = l; i <= r; i++) {
		if (q[i].op == 1) {
			if (tmp[i] + q[i].tmp >= q[i].k) q1[++l1] = q[i];
				else q[i].tmp += tmp[i], q2[++l2] = q[i];
		}
		else if (q[i].y <= mid) q1[++l1] = q[i];
			else q2[++l2] = q[i];
	}
	for (int i = l; i <= r; i++) {
		if (q[i].op == 0 && q[i].y <= mid) T.insert(q[i].x, -1);
		if (q[i].op == 2 && q[i].y <= mid) T.insert(q[i].x, 1);
	}
	for (int i = 1; i <= l1; i++) q[l + i - 1] = q1[i];
	for (int i = 1; i <= l2; i++) q[l + l1 + i - 1] = q2[i];
	
	work(l, l + l1 - 1, L, mid);
	work(l + l1, r, mid + 1, R);
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		q[++num] = (node){0, i, a[i], 0, 0, 0};
	}
	
	for (int i = 1; i <= m; i++) {
		op = getchar();
		while (op != 'C' && op != 'Q') op = getchar();
		if (op == 'Q') {
			scanf("%d %d %d", &x, &y, &k);
			q[++num] = (node){++tot, x, y, k, 1, 0};
		}
		else {
			scanf("%d %d", &x, &y);
			q[++num] = (node){0, x, a[x], 0, 2, 0};
			a[x] = y;
			q[++num] = (node){0, x, a[x], 0, 0, 0};
		}
	} 
	
	work(1, num, 0, 2e9);
	for (int i = 1; i <= tot; i++) printf("%d\n", ans[i]);
	
	return 0;
}