POJ 1328 Radar Installation(雷达安装)__贪心

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

分析

题目大意：
一共有n个小岛位于x轴之上, x轴为海岸, x轴上方为海洋，现需要在海岸上建立雷达, 每个雷达的覆盖半径为d，要求求出在海岸建立最少的雷达的数目, 使得可以覆盖所有的小岛。

首先雷达只能建在x轴上，而小岛可以为坐标系内任一点，计算是否可可覆盖小岛的方法是纪录下每个可覆盖小岛的圆的圆心的取值范围线段，如左端为(x - (d*d - y*d))，这样再将这些线段按左端点从小到大排序，每次雷达尽量布置在最右边，两个线段共同部分即是能覆盖这两个点的圆心取值范围。注意左、右端点要为double，我先开始就被这坑了没过。

实现

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 1010

struct Island
{
    double left, right;
}a[N];

bool operator <(Island &left, Island &right) {
    return left.left < right.left;
}

int solve(int n, int d)
{
    std::sort(a, a + n);   //让小岛的x坐标由小到大排列。
    int count = 1;
    double currentRight = a[0].right;
    for (int i = 1; i < n; i++) {
        if (a[i].left > currentRight) {
            count++;  //左端点是从小到大的，所以下一个线段与上一个线段范围不相交，则需要新加雷达站。
            currentRight = a[i].right;
        }
        else {
            //可布雷达站范围为相交线段(a[i].left, min(currentHeight, a[i].right)，布雷达站当然尽量往右布置。
            currentRight = currentRight > a[i].right ? a[i].right : currentRight;
        }
    }
    return count;
}

int main()
{
//  freopen("in.txt", "r", stdin);
    int n, d, x, y;
    int caseCount = 0;
    while (scanf("%d%d", &n, &d) && !(n == 0 && d == 0)) {
        double qrtD = d * d;
        int maxY = INT_MIN;
        for (int i = 0; i < n; i++) {
            scanf("%d%d", &x, &y);
            if (maxY < y) { maxY = y; }
            double delta = sqrt(qrtD - y * y);
            a[i].left = x - delta;   //以(x, 0)为圆心半径为d的圆投影到x轴上的线段是能覆盖的(x, y)点的雷达所能放置的范围。
            a[i].right = x + delta;
        }
        int result = (n == 0 && d != 0) ? 0 : (maxY > d ? -1 : solve(n, d));
        printf("Case %d: %d\n", ++caseCount, result);
    }
    return 0;
}