Poj 1328(greedy)

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 43100 Accepted: 9543

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

题意就是一个图上有n个点，要求你用尽量少的以在x轴上的点为圆心的圆覆盖这些点。对每一个点，要覆盖到它，圆心有一个确定的范围，那么对每一个点都处理出这些范围，然后问题就转化为了已知一些区间，要找最少的点，使这些区间中都至少存在一个点。这是个经典问题，对区间排序后，不停的对下一个区间取交集，如果交为空，ans++，更新当前区间范围。

#include<cstdio>#include<cstring>#include<string>#include<vector>#include<set>#include<map>#include<iostream>#include<algorithm>#include<queue>#include<cmath>using namespace std;typedef pair<int,int> P;const int maxn = 1000 + 5;struct Node{    double l,r;}a[maxn];bool cmp(Node a,Node b){    return a.l < b.l;}int main(){    int n,d;    int kase = 0;    while(scanf("%d%d",&n,&d) != EOF){        kase++;        if(n == 0 && d == 0) break;        int tag = 1;        for(int i = 0;i < n;i++){            int x,y;            scanf("%d%d",&x,&y);            if(y > d){                tag = 0;                continue;            }            double der = sqrt(d*d*1.0-y*y*1.0);            a[i].l = x-der;            a[i].r = x+der;        }        if(tag == 0){            printf("Case %d: -1\n",kase);            continue;        }        sort(a,a+n,cmp);        int ans = 1;        double nl = a[0].l,nr = a[0].r;        for(int i = 1;i < n;i++){            if(a[i].l < nr || fabs(a[i].l-nr) < 1e-6){                nr = min(nr,a[i].r);            }            else{                ans++;                nl = a[i].l;                nr = a[i].r;            }        }        printf("Case %d: %d\n",kase,ans);    }    return 0;}