POJ1328 Radar Installation 【貪心·區間選點】，poj1328radar

POJ1328 Radar Installation 【貪心·區間選點】，poj1328radar

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54593		Accepted: 12292

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

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>#define maxn 1010using namespace std;struct Node {    double u, v;    friend bool operator<(const Node& a, const Node& b) {        return a.u < b.u;    }} E[maxn];int N, D;int main() {    int i, ok, id, ans, cas = 1;    double x, y, d, flag;    while(scanf("%d%d", &N, &D), N) {        printf("Case %d: ", cas++);        ok = 1; id = 0;        for(i = 0; i < N; ++i) {            scanf("%lf%lf", &x, &y);            if(y > D) ok = 0;            if(!ok) continue;            d = sqrt(D * D - y * y);            E[id].u = x - d;            E[id++].v = x + d;        }        if(!ok) {            printf("-1\n");            continue;        }        sort(E, E + id);        flag = E[0].v; ans = 1;        for(i = 1; i < N; ++i) {            if(E[i].u <= flag) {                if(E[i].v <= flag) flag = E[i].v;                continue;            }            ++ans; flag = E[i].v;        }        printf("%d\n", ans);    }    return 0;}