poj 1328 Radar Installation【贪心】

Radar Installation

 

转载地址：http://blog.chinaunix.net/uid-22609852-id-3506134.html

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 53   Accepted Submission(s) : 27

Problem 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

分析：

这道题有些不懂，现在标记一下，往后来看。

代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#include<iostream>using namespace std;#define MAX 1010struct node {double xl;double xr;}is[MAX];int cmp(const void *a,const void *b){return ((node *)a)->xl>((node *)b)->xl?1:-1; }int gre(node is[],int n,int d){int num=0,i;double cur;cur=is[0].xr;num++;for(i=1;i<n;i++){if(is[i].xl-cur>1e-5){num++;cur = is[i].xr;}else {if(is[i].xr-cur<1e-5){cur=is[i].xr;}}}return num;}int main(){int i,n,d,x,y,cout=0,flag=0; node is[MAX]={0};double of;while(scanf("%d%d",&n,&d)==2&&!(n==d&&d==0)){flag=0;cout++;for(i=0;i<n;i++){scanf("%d %d",&x,&y);if(y>d){flag=1;}of=sqrt((double)(d*d-y*y));is[i].xl=x-of;is[i].xr=x+of;}if(flag){printf("Case %d: -1n",cout);continue;}qsort(is,n,sizeof(is[0]),cmp);i=gre(is,n,d);printf("Case %d: %d",cout,i);} return 0;}