Poj1328-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

 

 真想说句，“啥也不想说了！！！”，qsort函数排序，人家从零开始，我赋值偏要从‘1’开始，调了半天

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
const int MAXN = 10000 + 10;

#define FFF freopen("input.txt", "r", stdin)

struct Node
{
    double x, y;
} node[MAXN];

int cmp(const void *a, const void *b)
{
    return (*(Node *)a).x > (*(Node *)b).x ? 1 : -1;
}

int main()
{
    FFF;
    int n;
    double r;
    int cas = 1;
    while(scanf("%d %lf", &n, &r) != EOF && (n+r))
    {
        bool tag = false;
        for(int i = 0; i < n; ++i)
        {
            scanf("%lf %lf", &node[i].x, &node[i].y);
            if(node[i].y > r)
            {
                tag = true;
            }
        }
        if(tag)
        {
            printf("Case %d: ", cas++);
            printf("-1\n");
            continue;
        }
        ///给他进行排序，然后从一端进行扫描
        qsort(node, n, sizeof(node[0]), cmp);///qsort人家是从下标为0，进行排序，以后注意！！

        int sum = 1;

        printf("Case %d: ", cas++);

        double left[MAXN],righ[MAXN];    //求出每个小岛与坐标轴的交点
        for(int i=0 ; i < n; i++)
        {
            left[i] = node[i].x - sqrt(r*r - node[i].y * node[i].y);
            righ[i] = node[i].x + sqrt(r*r - node[i].y * node[i].y);
        }

        double temp = righ[0];      //将第一个雷达放置在第一个点与坐标轴的右交点
        for(int i = 0; i < n-1; ++i)
        {
            if(left[i+1] > temp)    //这种情况下，不论雷达怎么放置，两小岛都不能在同一雷达范围内，所以必须放置新的雷达
            {
                temp = righ[i+1];
                sum++;
            }
            else if(righ[i+1] < temp)//当在这种情况下，必须要将雷达所在的位置更新为新小岛与坐标轴的右边界，才能保证
            {
                temp = righ[i+1];    //雷达能够覆盖到这两个岛屿
            }                        //在其他情况就，不用放置新的雷达，也不用更新雷达的坐标
        }
        printf("%d\n", sum);
    }
    return 0;
}