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 2 1 2 -3 1 2 1 1 2 0 2 0 0 Sample Output Case 1: 2 Case 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; }