B. 【例题2】雷达装置

$$B.【例题2】雷达装置$$

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题目解析

求最少所需的雷达数,考虑贪心算法.

以这张图为例.以一个城市为中心,作一个半径为$d$的圆.而这个圆与$x$轴的左交点$l$到右交点$r$的区间$[l,r]$内的说有点放置雷达都可以至少笼罩住这个城市.

因为是贪心思想,那么就尽量是最多区间的点,并且优先最右端.

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Code

#include <bits/stdc++.h>
#define db double
using namespace std;

struct abc{
	db x, y;
}a[1005];
db h[10005];
int n, d, ans;

bool cmp(abc a, abc b)
{
	return a.y < b.y;
}

int main()
{
	scanf ("%d%d", &n, &d);
	for (int i = 1; i <= n; ++ i)
	{
		int t1, t2;
		scanf ("%d%d", &t1, &t2);
		db er = sqrt (d * d - t2 * t2);
		if(t2 > d)
		{
			ans = -1;
			break;
		}
		a[i].x = t1 - er;
		a[i].y = t1 + er;
	}
	if (ans < 0) 
	{
		printf ("-1");
		return 0;
	}
	sort (a + 1, a + n + 1, cmp);
	h[ ++ ans] = a[1].y;
	for (int i = 2; i <= n; ++ i)
	{
		if (h[ans] < a[i].x)
			h[ ++ ans] = a[i].y;
	}
	printf ("%d", ans);
	return 0; 
}