radar（nyoj287贪心）

Radar

时间限制：1000 ms  |  内存限制：65535 KB

难度：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.

 

输入

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

输出

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.

样例输入

3 21 2-3 12 11 20 20 0

样例输出

Case 1: 2Case 2: 1

#include<stdio.h>#include<math.h>#include<stdlib.h>#define MAX 1005int num = 1;struct land{double l;double r;}land_1[MAX];int cmp(const void *a, const void *b)                          //  在c中运用qsort函数进行排序所需的自己的cmp函数标准模式{return (*(struct land *) a). r > (*(struct land *) b). r ? 1 : -1;//从小到大进行排序}int main(){int count, r, x, y, flag, num_1, i;double now, result;while(scanf("%d%d", &count, &r) && count && r){flag = 1;for(i = 0; i < count; i++){scanf("%d%d", &x, &y);if(!flag)continue;if(y <= r){result =sqrt((double) (r * r -  y * y));land_1[i].l = (double)x - result;land_1[i].r = (double)x + result;//flag = 1;}elseflag = 0;if(!flag){printf("Case %d: -1\n", num++);continue;}}qsort(land_1, count, sizeof(land_1[0]), cmp);//qsort函数的用法num_1 = 1;now = land_1[0].r;for(i = 1; i < count; i++){if(land_1[i].l <= now + 0.00005)continue;now = land_1[i]. r;num_1++;}printf("Case %d: %d\n", num++, num_1);}return 0;}