POJ 1328 Radar Installation 贪心

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 66208 Accepted: 14832

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

和以前做的一道一样Orz

ACcode：

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 100005#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;struct me{    double f,s;}my[1111];bool comp(me a,me b){    return a.s<b.s;}int main(){    int n,r,i;double x,y;    int ccc=1;    while(rd2(n,r)!=EOF&&n&&r){        bool flag=false;        for(i=0;i<n;++i){            cin>>x>>y;            if(y>r)            flag=true;            my[i].f=x-sqrt(r*r-y*y);            my[i].s=x+sqrt(r*r-y*y);        }        sort(my,my+n,comp);        int cns=1;        double tt=my[0].s;        for(i=0;i<n;++i){            if(tt<my[i].f){                cns++;                tt=my[i].s;            }        }        printf("Case %d: ",ccc++);        printf(flag?"-1\n":"%d\n",cns);    }    return 0;}/*3 21 2-3 12 11 20 20 0*/