poj 1328 Radar Installation
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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 2
1 2
-3 1
2 1
1 2
0 2
0 0
Sample Output
Case 1: 2
Case 2: 1
思路:算出 每个小岛能被覆盖的雷达的圆心,即以小岛为圆心 R为半径 作圆,该圆与X轴的交点:左交点为x-sqrt(R*R-y*y); 右交点为x+sqrt(R*R-y*y);按照 左交点 排序,如果i点的左交点在 当前雷达的右边 则需安装一个新雷达,更新雷达否则 如果 i点的右交点也在当前雷达的左边 则把当前雷达的圆心更新为该点的右交点;
#include<stdio.h>#include<iostream>#include<cmath>#include<string.h>#include<algorithm>using namespace std;struct node{ double l,r;} p[1000];bool cmp(node a,node b){ return a.l<b.l;}int main(){ int n,d,i,x,y,sw,re,count = 1; double pre; while(1) { scanf("%d %d",&n,&d); if(n == 0 && d==0) break; sw = 1; for(i=0; i<n; i++) { scanf("%d %d",&x,&y); if(d>=y&&sw==1) { p[i].l = x-sqrt((double)d*d - (double)y*y); p[i].r = x+sqrt((double)d*d - (double)y*y); } else { sw = 0; } } if(sw == 0) { printf("Case %d: -1\n",count++); continue; } sort(p,p+n,cmp); re = 1; pre = p[0].r; for(i=1; i<n; i++) { if(p[i].l>pre) { re++; pre = p[i].r; } else { if(p[i].r<pre) { pre = p[i].r; } } } printf("Case %d: %d\n",count++,re); } return 0;}
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