poj 1328 Radar Installation 排序贪心

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 56702		Accepted: 12792

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
思路就是，排序之后，按右端点放雷达就好

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>



using namespace std;



const int N=1010;



struct node{

    double l,r;

}seg[N];



int cmp(node a,node b){

    return a.l<b.l;

}



int main(){



    //freopen("input.txt","r",stdin);



    int n,d,cases=0;

    while(~scanf("%d%d",&n,&d)){

        if(n==0 && d==0)

            break;

        int x,y;

        double tmp;

        int flag=0;

        for(int i=0;i<n;i++){

            scanf("%d%d",&x,&y);

            tmp=sqrt((double)(d*d)-y*y);

            seg[i].l=x-tmp;

            seg[i].r=x+tmp;

            if(y>d)

                flag=1;

        }

        sort(seg,seg+n,cmp);

        printf("Case %d: ",++cases);

        if(flag){

            printf("-1\n");

            continue;

        }

        int ans=1;

        node line=seg[0];

        for(int i=1;i<n;i++){

            if(line.r<seg[i].l){

                ans++;

                line=seg[i];

            }

            else  if(line.r>=seg[i].r)

                line=seg[i];



        }

        printf("%d\n",ans);

    }

    return 0;

}