Poj 1068 Radar Installation(模拟水题)

题目

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

 

题解

此题思路不是问题，主要在与需要对结构体的某一元素进行排序的时候，sort调用的方法。

 例如：

struct A
{
       int a;
       int b;
};
A arr[10];
//想要实现对arr中每个元素的a成员排序

bool cmp(A &a,A &b)
{
     return a.a < b.a;
}

sort(arr, arr + n,cmp);  //即可

 

代码

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
double cal(int x, int y, int r)
{
    return (double)x+pow((r*r - y*y),0.5);
}
struct point
{
    int x, y;
    double r;
};
bool comp(point &p1, point &p2)
{
    return p1.r < p2.r;
}
bool ok(int x, int y, double r,int R)
{
    return ((double)x - r)*((double)x - r) + (double)y*(double)y <= (double)R*(double)R;
}
int main()
{
    int q = 1;
    while (true)
    {
        int N, R;
        int n = 0;
        point p[1000]; 
        int flag[1000];
        scanf("%d %d", &N, &R);
        if (N == 0 && R == 0) break;
        for (int i = 0; i < N; i++)
        {
            scanf("%d %d", &p[i].x, &p[i].y); flag[i] = 0;
            if (p[i].y > R) n = -1;
            p[i].r = cal(p[i].x, p[i].y, R);
        }
        if (n != -1) {
        sort(p, p + N, comp);
        double cur;
        for (int i = 0; i < N; i++)
        {
            if (flag[i] == 0)
            {
                n++;
                cur = p[i].r;
                if (i == N - 1) break;
                for (int j = i + 1; j < N; j++)
                {
                    if (p[j].r - cur > 2 * R) break;
                    if (ok(p[j].x, p[j].y, cur, R)) flag[j] = 1;
                }
            }
        }
        }
        printf("Case %d: %d\n",q, n);
        q++;
    }
}