Sicily 13860. Crowded Cows

13860. Crowded Cows

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

Farmer John's N cows (1 <= N <= 50,000) are grazing along a one-dimensional fence. Cow i is standing at location x(i) and has height h(i) (1 <= x(i),h(i) <= 1,000,000,000).

A cow feels "crowded" if there is another cow at least twice her height within distance D on her left, and also another cow at least twice her height within distance D on her right (1 <= D <= 1,000,000,000). Since crowded cows produce less milk, Farmer John would like to count the number of such cows. Please help him.

Input

* Line 1: Two integers, N and D.

* Lines 2..1+N: Line i+1 contains the integers x(i) and h(i). The locations of all N cows are distinct.

Output

* Line 1: The number of crowded cows.

Sample Input

6 4
10 3
6 2
5 3
9 7
3 6
11 2

Sample Output

2

Hint

In the sample, there are 6 cows, with a distance threshold of 4 for feeling crowded. Cow #1 lives at position x=10 and has height h=3, and so on. The cows at positions x=5 and x=6 are both crowded.

Problem Source

2015年每周一赛第二场

按x从小到大排序，而后从左往右扫，维护一个x递增，h递减的双头队列。

对于新的牛，从队列左侧开始，弹出位置超过限制的牛，这些牛必然比新牛及后来的牛位置更远，从队列右侧开始，弹出高度没有新牛高的牛，因为这些牛比新牛矮而且位置也

更远，都是必然没有新牛有可能符合“拥挤”的牛。

这是确定左边拥挤，同理确定右边拥挤即可。

#include 
#include 
using namespace std;

struct cow {
	int x, h;
}c[50005];

bool isLeftCrowded[50005];

bool cmp(const cow & c1, const cow & c2) {
	return c1.x < c2.x;
}

int main() {
	std::ios::sync_with_stdio(false);
	int n, d;
	cin >> n >> d;
	for (int i = 0; i < n; i++) cin >> c[i].x >> c[i].h;
	sort(c, c + n, cmp);
	int q[50005];
	int head, tail, ans = 0;
	head = tail = 0;
	q[0] = 0;
	for (int i = 1; i < n; i++) {
		int nearestPos = c[i].x - d;
		while (head <= tail && c[q[head]].x < nearestPos) head++;
		while (head <= tail && c[q[tail]].h <= c[i].h) tail--;
		if (head > tail) q[head] = i;
		if (c[q[head]].h >= c[i].h << 1) isLeftCrowded[i] = true;
		q[++tail] = i;
	}
	head = tail = 0;
	q[0] = n - 1;
	for (int i = n - 2; i >= 0; i--) {
		int nearestPos = c[i].x + d;
		while (head <= tail && c[q[head]].x > nearestPos) head++;
		while (head <= tail && c[q[tail]].h <= c[i].h) tail--;
		if (head > tail) q[head] = i;
		if (c[q[head]].h >= c[i].h << 1 && isLeftCrowded[i]) ans++;
		q[++tail] = i;
	}
	cout << ans << endl;
	return 0;
}