Radar Installation

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Radar Installation

Time Limit:1000MS Memory Limit:10000K
Total Submit:55 Accepted:18

Description

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.

Figure A Sample Input of Radar Installations

Input

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

Output

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.

Sample Input

3 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

#include <iostream>#include <cmath>using namespace std;int x[1010], y[1010];struct PN{double left;double right;}q[1010];int cmp( const void *a, const void *b ){PN *c = (PN *)a;PN *d = (PN *)b;if( c->left>d->left )///////////升序return 1;return -1;}int main(){int i, n, d, sign, s=0;while( scanf("%d%d", &n, &d)!=EOF&& (n||d) ){sign = 1;if( d<=0 )sign = 0;for( i=1; i<=n; ++i ){scanf("%d%d", &x[i], &y[i]);if( y[i]>d&&sign )sign = 0;}s++;printf("Case %d: ",s);if( sign ){for( i=1; i<=n; ++i ){double t = sqrt(double(d*d-y[i]*y[i]));q[i].left = x[i]-t;q[i].right = x[i]+t;}qsort(q+1,n,sizeof(q[1]),cmp);int min;double  high;min=1;high = q[1].right;for( i=2; i<=n; ++i ){if( q[i].left>high ){min++;high = q[i].right;}elsehigh = q[i].right<high?q[i].right:high;}printf("%d/n", min);}else{printf("-1/n");}}return 0;}

以区间左端点升序排列，选择右端点为圆心。每次查找是否能将下一个点包含进来。