poj 1328 Radar Installation (贪心）@

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：在x轴上方，分布一些点，求在x轴上最少放置几个雷达可以将他们全部包围，给定雷达的检测范围；

解：从左到右贪心，预处理出一个点可以被包含的区间长度，求更新一个区间的长度；

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>using namespace std;const int N = 1e6+10;typedef long long LL;const LL mod = 1e9+7;const double eps= 1e-6;struct node{    double l, r;}p[N];int cmp(node x,node y){    if(x.l==y.l) return x.r<y.r;    return x.l<y.l;}int main(){    int n, ncase=1;    double d;    while(scanf("%d %lf", &n, &d),n!=0||d!=0)    {        int flag=0;        for(int i=0;i<n;i++)        {            double x, y;            scanf("%lf %lf", &x, &y);            if(y>d||y<0) flag=1;            p[i].l=x-sqrt(d*d-y*y), p[i].r=x+sqrt(d*d-y*y);        }        sort(p,p+n,cmp);        int cnt=1;        double l=p[0].l, r=p[0].r;        for(int i=1;i<n;i++)        {            if(p[i].l>r)            {                cnt++;                l=p[i].l, r=p[i].r;            }            else if(p[i].r<=r)  r=p[i].r;        }        if(flag) printf("Case %d: -1\n",ncase++);        else printf("Case %d: %d\n",ncase++,cnt);    }    return 0;}