poj 1328 Radar Installation （贪心）

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本文介绍了一个经典的算法问题——如何使用最少数量的雷达装置覆盖海上的所有小岛。通过计算每个小岛可以被雷达覆盖的范围，并采用贪心算法确定雷达的最佳放置位置。

链接:poj 1328

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

Sample Output

Case 1: 2
Case 2: 1

题意：X轴表示海岸线，在X轴上方是海，下方是陆地，海上有n个小岛，现在要在海岸线上安装雷达，

雷达的覆盖范围是一个以雷达为圆心，以d为半径的圆，请用最少的雷达覆盖所有的小岛；

当无法覆盖时输出-1

思路：算出每个小岛能被覆盖的雷达的圆心范围，即以小岛为圆心 d为半径作圆，该圆与X轴的交点：

左交点为x-sqrt(d*d-y*y);

右交点为x+sqrt(d*d-y*y);

按照左交点从小到大排序，第一个雷达放在第一个点的右交点，

如果某点的左交点在当前雷达的右边，则需安装一个新雷达，更新雷达的位置

否则如果某点的右交点也在当前雷达的左边，则把当前雷达的圆心更新为该点的右交点；

贪心思想，开始wrong了好多遍，同样的思路，网上解题报告代码一遍就AC,始终想不明白，突然发现记录雷达的坐标定义成了 int型，改了就A了，以后得细心、、、

AC代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct stu
{
    double x,y,l,r;
};
int cmp(struct stu a,struct stu b)
{
    return a.l<b.l;
}
int main()
{
    int i,j,n,d,m,t;
    double k;
    struct stu s[1005];
    for(i=1;;i++){
        scanf("%d%d",&n,&d);
        t=1;
        if(n==0&&d==0)
            break;
        for(j=0;j<n;j++){
            scanf("%lf%lf",&s[j].x,&s[j].y);
            s[j].l=s[j].x-sqrt(d*d-s[j].y*s[j].y);
            s[j].r=s[j].x+sqrt(d*d-s[j].y*s[j].y);
            if(s[j].y>d||d<=0)
                t=0;
        }
        if(t==0){
            printf("Case %d: -1\n",i);
            continue;
        }
        sort(s,s+n,cmp);
        k=s[0].r;
        m=1;
        for(j=1;j<n;j++){
            if(s[j].l>k){
                m++;
                k=s[j].r;
            }
            else if(s[j].r<k)
                k=s[j].r;
        }
        printf("Case %d: %d\n",i,m);
    }
    return 0;
}