radar

Radar

時間限制：1000 ms  |  內存限制：65535 KB

難度：3

描述

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d. 

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

 

輸入

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 

The input is terminated by a line containing pair of zeros

輸出

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

樣例輸入

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

樣例輸出

Case 1: 2
Case 2: 1

第一種方案：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define M 10000

typedef struct data
{
	double left, right;
}Data;

int com(const void *p, const void *q)
{
	return ((Data *)p)->left > ((Data *)q)->left ? 1 : -1;
}

int main()
{
	double n, m;
	int num = 0, count = 0;
	int i = 0, result = 0, mark = 0;	
	Data data[M];
	double a[M][2], r = 0, me;

	for(;;)
	{
		num = 0;
		mark = 0;	
		memset(a, -1, 2 * M);	
		scanf("%lf %lf", &n, &m);
		getchar();

		if(n == 0 && m == 0)
			return 0;
		for(i = 0; i < n; i++)
		{
			scanf("%lf %lf", &a[i][0], &a[i][1]);
			getchar();
			if(a[i][1] > m)//如果島嶼的位置大於雷達的輻射半徑，說明不可能有解決方案
				mark = 1;				
		}	
		for(i = 0; i < n; i++)
		{
			r = m * m - a[i][1] * a[i][1];
			data[i].left = a[i][0] - sqrt(r);
			data[i].right = a[i][0] + sqrt(r);
		}
		qsort(data, n, sizeof(Data), com);//按雷達輻射範圍的左邊座標從左到右排序

		me = data[0].right;//將第一個雷達放到第一個島嶼的右座標上
		num ++;//次數加一
		for(i = 1; i < n; i++)//從第二個島嶼的範圍開始遍歷
		{
			if(data[i].left > me + 0.00005)//如果下一個島嶼的左座標大於當前的雷達說明雷達要再加一個且放到下一個島嶼的右座標，並將其標記
			{
				me = data[i].right;
				num++;
			}
			else //如果下一個島嶼的右座標還小與當前的雷達位置，說明當前雷達輻射不到下一個島嶼，只要將當前雷達移到下一個島嶼的右座標就可以了
			{	
				if(data[i].right < me + 0.00005)		
					me = data[i].right;
			}
		}
		if(mark == 0)
			printf("Case %d: %d\n", ++count, num);
		else
			printf("Case %d: -1\n", ++count);
	}

	return 0;
}

第二種方案：

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#define M 1000

typedef struct data
{
	double x, y;
}Data;

int count = 0;
double n = 0, m = 0;
int find(Data data[]);
int com(const void *p, const void *q);

int main()
{
	int i = 0, result = 0;
	Data data[M];

	for(;;)
	{		
		scanf("%lf %lf", &n, &m);
		getchar();

		if(n == 0 && m == 0)
			return 0;
		for(i = 0; i < n; i++)
		{
			scanf("%lf %lf", &data[i].x, &data[i].y);
			getchar();					
		}	
		qsort(data, n, sizeof(Data), com);//按照雷達的位置從左到右排序
		result = find(data);
		printf("Case %d: %d\n", ++count, result);
	}
	return 0;
}

int com(const void *p, const void *q)
{
	return ((Data *)p)->x - ((Data *)q)->x;
}

int find(Data data[])
{
	int i, num = 0;
	double me;
	double x[2] = {0}, r;
	
	if(m < data[0].y)
	{
		return -1;
	}
	r = m * m - data[0].y * data[0].y;
	x[1] = data[0].x + sqrt(r);
	me = x[1];
	num++;
	for(i = 1; i < n; i++)
	{
		if(m < data[i].y || data[i].y < 0)//如果島嶼的位置小於0或者大於雷達的輻射半徑，則說明沒有解決方案
		{
			num = -1;
			break;
		}
		r = m * m - data[i].y * data[i].y;
		x[0] = data[i].x - sqrt(r);
		x[1] = data[i].x + sqrt(r);
		if(me < x[0]) //如果當前雷達的位置小於下一個島嶼的左座標則需要再添加一個雷達且添加位置爲下一個島嶼的右座標
		{
			me = x[1];
			num++;
		}
		else 
		{
			if(x[1] < me)//如果下一個島嶼的右座標小於當前雷達，說明當前的雷達輻射不到下一個島嶼，則將當前雷達移到下一個島嶼的右座標
				me = x[1];
		}
 	}
	return num;
}