poj 1328 Radar Installation(贪心)

Radar Installation

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Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 55857 Accepted: 12596

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

Source

Beijing 2002

题目大意：在直角坐标系中，x轴为海岸，x轴上方为海（岛屿在x轴上方），下方为陆地，告诉你岛屿的坐标，以及雷达的探测半径d，求最小数量的雷达，将岛屿覆盖，如果不能则输出-1；

显然在d<y（岛屿的纵坐标）时，不可能有雷达覆盖他，这是输出-1；

此题采用贪心策略： 以岛屿i坐标(p[i].x,p[i].y)为圆心，做半径为d的圆，交x轴两个交点，p[i].l记为左交点，p[i].r记为右交点，将岛屿按照p[i].l排序，按照排序后从左到右依次选择圆心所在位置（圆心可以不为整数），使得该圆能够覆盖足够多的点，首先选择第一个点的p[i].r作为圆心记为cr，然后判断，当（p[i].l>cr）时，显然，当前的圆覆盖不了此点，只能新建一个，ans++，否则当p[i].r<cr时，说明还有更优的选择，将当前cr更新为p[i].r；

此外此题还应注意一点：用qsort排序时，下面这样会导致精度丢失，导致WA

int cmp(const void *a,const void *b){Point *p1=(Point *)a,*p2=(Point *)b;if(p1->l!=p2->l)return p1->l-p2->l;return p1->r-p2->r;}

应该这样写：

int cmp(const void *a,const void *b){Point *p1=(Point *)a,*p2=(Point *)b;if(p1->l>p2->l)return 1;if(p1->l<p2->l)return -1;if(p1->r>p2->r)return 1;if(p1->r<p2->r)return -1;return 0;}

代码：

#include <iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<stdlib.h>#define N 1010#define getx2(x) (x)*(x)#define eps 1e-10/* run this program using the console pauser or add your own getch, system("pause") or input loop */using namespace std;struct Point{double l,r;int x,y;}p[N];void init(int n){for(int i=0;i<=n;i++){p[i].l=p[i].r=p[i].x=p[i].y=0;}}int cmp(const void *a,const void *b){Point *p1=(Point *)a,*p2=(Point *)b;if(p1->l>p2->l)return 1;if(p1->l<p2->l)return -1;if(p1->r>p2->r)return 1;if(p1->r<p2->r)return -1;return 0;}int main(int argc, char** argv) {int n,d,cnt=0,i;bool flag;while(~scanf("%d%d",&n,&d)&&(n||d)){flag=true;init(n);     for(i=0;i<n;i++)     {     scanf("%d%d",&p[i].x,&p[i].y);     if(p[i].y>d)     {     flag=false;     continue; }     p[i].l=p[i].x-sqrt(getx2(d)-getx2(p[i].y));     p[i].r=p[i].x+sqrt(getx2(d)-getx2(p[i].y)); } if(!flag) { printf("Case %d: -1\n",++cnt); continue; } qsort(p,n,sizeof(p[0]),cmp); int ans=1; double cr=p[0].r; for(i=1;i<n;i++) { if(p[i].l>cr) { ans++,cr=p[i].r; } else if(p[i].r<cr)  cr=p[i].r; } printf("Case %d: %d\n",++cnt,ans);}return 0;}