poj1328--Radar Installation
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Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 49680 Accepted: 11104
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
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
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
Source
Beijing 2002
小贪心,由左向右遍历一次,对每一个点找出对应的右侧的圆心,由左向右,确定圆心后,判断之后的点会不会包含在已经确定的圆中
#include <stdio.h>#include <math.h>#include <algorithm>using namespace std;struct node{ int x , y ; double s ;} p[2000];bool cmp(node a,node b){ return a.s < b.s ;}int main(){ int n , b , i , k = 0 ; while(scanf("%d %d", &n, &b)!=EOF) { if(n == 0 && b == 0) break; int num = 0 ; double x ; k++ ; int flag = 1 ; for(i = 0 ; i < n ; i++) { scanf("%d %d", &p[i].x, &p[i].y); if(p[i].y > b) flag = 0 ; else p[i].s = p[i].x * 1.0 + sqrt(b*b*1.0 - p[i].y *p[i].y *1.0) ; } if(flag == 0) { printf("Case %d: -1\n", k); continue ; } sort(p,p+n,cmp); x = p[0].s ; num = 1 ; for(i = 1 ; i < n ; i ++) { if( sqrt( p[i].y*p[i].y*1.0 + (x-p[i].x) * (x-p[i].x) *1.0 ) > b ) { num++ ; x = p[i].s ; } } printf("Case %d: %d\n", k, num); } return 0 ;}
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