POJ 1328 Radar Installation 雷達安裝 貪心問題求解

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題目連結： POJ 1328 Radar Installation

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

題意

在x軸表示的海岸線上需要布置雷達，以覆蓋海中的n個海島，雷達的覆蓋半徑為d，求解如何選擇布置點才能使用到的雷達最少。如果有的海島不能被覆蓋到，那麼輸出-1。

分析

考察以海島為圓心，做半徑為d的圓，看與x軸相交的那段區間，這樣的話，這個區間內的任何位置布置雷達，都是可以覆蓋這個海島的，對於所有的海島，當然不乏求到的區間有部分重合的情況，那麼在這個重合的區間中布置雷達，當然就能覆蓋到兩個以上的點，這樣就能節省雷達，雷達所在的區間越多，節省的雷達就越多。

思想

典型的貪心思想。對於求出的每一個區間，我們進行排序，讓區間右端點小的排在前面，如果右端點相等，那麼左端點大的排在前面（想想為什麼）。

那麼該如何選擇布置點呢？

首先我們選取排序號後的第一個區間的右端點為第一個布置點，它的位置為st，然後再按順序找後面的區間。如果當前區間的左端點大於st，說明st位置的雷達不能覆蓋到這個海島，那麼雷達數加1，同時更新st為這個區間的右端點。如果當前區間的左端點小於等於st，則說明st位置的雷達能覆蓋這個海島，忽略這個區間。

代碼

/*    POJ_1328_Radar_Installation    Author: Sign_    Greedy*/#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;int n, d;struct seg{    double l, r;}SEG[1010];seg pos(int x, int y){    seg s;    s.l = x - sqrt(d*d - y*y);    s.r = x + sqrt(d*d - y*y);    return s;}bool cmp(seg a, seg b){    if(a.r == b.r) return a.l > b.l;    return a.r < b.r;}int main(){    int x[1010], y[1010], cas = 1;    while(scanf("%d%d", &n, &d), n, d)    {        bool flag = true;        for(int i = 0; i < n; i++)        {            scanf("%d%d", &x[i], &y[i]);            if(y[i] > d)                flag = false;        }        if(!flag)        {            printf("Case %d: -1\n", cas++);            continue;        }        for(int i = 0; i < n; i++)            SEG[i] = pos(x[i], y[i]);        sort(SEG, SEG+n, cmp);        int cnt = 1;        double st = SEG[0].r;        for(int i = 1; i < n; i++)            if(SEG[i].l > st)            {                st = SEG[i].r;                cnt++;            }        printf("Case %d: %d\n", cas++, cnt);    }    return 0;}