POJ_1328_Radar Installation(greedy)

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 59449 Accepted: 13394

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

题意：二维坐标系中有n个点，现在要你用最少的圆（圆心在x轴，半径为d）覆盖所有的点。

分析：贪心题。先预处理出每个点对应的圆心坐标区间，意思就是在这个区间内的任意一个点作为圆心坐标都能覆盖该点。然后根据区间右边界升序排序。然后就是区间有无交集的问题了。如果有交集，说明可以用同一个圆覆盖，如果没交集，ans+=1。

题目链接：http://poj.org/problem?id=1328

#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<cctype>#include<string>#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;const int maxn = 1000 + 5;struct Point{    double x;    double y;}point[maxn];int n,k=1;double d,X,Y;bool judge;bool cmp(Point a,Point b){    return a.y<b.y;}void input(){    judge=true;    for(int i=0;i<n;i++){        scanf("%lf%lf",&X,&Y);        if(fabs(Y)>d || d<=0){            judge=false;        }        else{            point[i].x=X-sqrt(d*d-Y*Y);            point[i].y=X+sqrt(d*d-Y*Y);        }    }}void solve(){    printf("Case %d: ",k++);    if(!judge || d<=0){        printf("-1\n");        return ;    }    sort(point,point+n,cmp);    int ans=1;    double maxy=point[0].y;    for(int i=1;i<n;i++){        if(point[i].x>maxy){            maxy=point[i].y;            ans++;        }    }    printf("%d\n",ans);    return ;}int main(){    while(scanf("%d%lf",&n,&d)!=EOF){        if(n==0&&d==0) break;        input();        solve();    }return 0;}