nyoj 题目287 Radar 贪心算法

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.

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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 2
1 2
-3 1
2 1

1 2
0 2

0 0

样例输出

Case 1: 2
Case 2: 1

来源

Beijing 2002

上传者

ctest 大意： 海上有n个海岛，图中y上半轴为海，下半轴为陆地。现布置一些雷达在x轴上，作用半径为d,求覆盖全部海岛的最少雷达数。输入第一行为n,d;以后的n行为n个海岛的坐标x,y；以0，0代表结束，如果不能完全覆盖，输出-1. 思路： 如果雷达可以覆盖到海岛，那么以海岛为圆心，以d为半径画圆在x轴必相交（最坏也是相切），那么就有一段区间，在这个区间内雷达在哪都能覆盖这个海岛，如果两个海岛作圆，如果区间有重叠，在重叠处放置一个雷达就可覆盖两个海岛，以此类推。那么就转化成了贪心算法中区间找点的问题，将每个海岛在x轴形成的坐标区间，保存到数组，并排序，再判断。

#include
#include
#include
struct fun
{
double top;//左坐标 
   double last;//右坐标 
}s[1000];
int  cmp(const void*a,const void*b)
{
return (*(fun*)a).last>(*(fun*)b).last?1:-1;
}
int main()
{
int k=0;
double n,r;
while(~scanf("%lf %lf",&n,&r))
{
int flag=1;
if(n==0&&r==0) break;
double x,y;
for(int i=0;ir)//海岛距离太远，雷达半径不够，不能覆盖 
{
flag=0;
break;
}
s[i].top = x-sqrt(r*r-y*y); 
s[i].last = x+sqrt(r*r-y*y);
}
qsort(s,n,sizeof(s[0]),cmp);//以右坐标从小到大排序 
int sum=1;
double temp=s[0].last ;//左边第一个肯定放置一个，
 for(int i=1;i temp)//左坐标大于上一个右坐标 ，不重叠，要放置雷达 
{
sum++;
temp=s[i].last ;
}

}
if(flag==1)
printf("Case %d: %d\n",++k,sum);
else printf("-1\n");
}
return 0;
}