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 2
1 2
-3 1
2 1

1 2
0 2

0 0
Sample Output

Case 1: 2
Case 2: 1
Source

Beijing 2002

求每个岛对应的覆盖它的雷达的有效区间，然后扫描区间去掉重合的。
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
struct Island
{
    int x; // 岛屿x坐标
    int y; // 岛屿y坐标
    double minRadarX; // 覆盖它的雷达有效区间的左端
    double maxRadarX; // 覆盖它的雷达有效区间的右端
};

bool IslandCmp(Island & island1, Island & island2)
{
    return island1.x < island2.x;
}
int main()
{
    int n, d;
    int caseNo = 0;
    vector<Island> islandsVec;
    while(cin >> n >> d)
    {
        islandsVec.clear();
        if(n == 0 || d == 0)
            break;
        caseNo++;
        Island island;
        for(int i = 0; i < n; i++)
        {
            cin >> island.x >> island.y;
            islandsVec.push_back(island);
        }

        // 按岛屿x坐标排序
        sort(islandsVec.begin(), islandsVec.end(), IslandCmp);

        // 求每个岛屿对应的雷达的有效区间
        bool bNoAnswer = false;
        vector<Island>::iterator iter;
        for(iter = islandsVec.begin(); iter != islandsVec.end(); iter++)
        {
            if(iter->y > d)
            {
                bNoAnswer = true;
                break;
            }
            double r = sqrt(d * d * 1.0 - iter->y * iter->y);
            iter->minRadarX = iter->x - r;
            iter->maxRadarX = iter->x + r;
        }

        if(bNoAnswer)
        {
            cout << "Case " << caseNo << ": " << -1 << endl;
        }
        else
        {
            // 统计实际需要的雷达数目
            int result = 0;
            double preEnd;
            vector<Island>::iterator iter;
            for(iter = islandsVec.begin(); iter != islandsVec.end(); iter++)
            {
                if(iter == islandsVec.begin())
                {
                    result++;
                    preEnd = iter->maxRadarX;
                }
                else
                {
                    if(iter->minRadarX > preEnd)
                    {
                        result++;
                        preEnd = iter->maxRadarX;
                    }
          &nbs

补充：软件开发 , C++ ,