POJ 1328 Radar Installation

Radar Installation

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1328

Appoint description:   System Crawler  (2015-04-04)

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

题意  ：有n个小岛 并给出海岸线上雷达覆盖的半径d，雷达只能放置于海岸线上（x轴），给出每个小岛的坐标，问至少要多少个雷达才能把小岛全部覆盖。

解法一：（左交点 从小到大排序）

      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（因为：

有这种情况。）

  

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
using namespace std;
struct node
{
    double left, right;
}island[1001];
bool cmp(node a, node b)
{
    return a.left < b.left;
}
int main()
{
    double x, y, d, temp;
    int i, cnt, n, k = 0;
    bool flag;
    while (scanf("%d%lf", &n, &d) && !(n == 0 && d == 0))
    {
        k++;
        flag = false;
        for (i = 0; i<n; i++)
        {
            scanf("%lf%lf", &x, &y);
            if (y > d || d<0)
            {
                flag = true;
            }
            island[i].right = x + sqrt(d*d - y * y);
            island[i].left = x - sqrt(d*d - y * y);
        }
        if (flag)
        {
            printf("Case %d: -1
", k);
            continue;
        }
        sort(island, island + n, cmp);
        temp = island[0].right;
        cnt = 1;
        for (i = 1; i<n; i++)
        {//分两种
            if (island[i].right <= temp)
            {
                temp = island[i].right;
            }
            else if (island[i].left > temp)
            {
                cnt++;
                temp = island[i].right;
            }
        }
        printf("Case %d: %d
", k, cnt);
    }
    return 0;
}

 解法二：排序按右端点排，右小的在前，再左大的在前

则只需考虑 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

#include<iostream>  
#include<cstdio>  
#include<string> 
#include<cstring>  
#include<algorithm>  
#include<cmath>  
using namespace std;
struct ac
{
    double l;
    double r;
}p[1500];

bool cmp(ac x, ac y)//排序按右端点排，右小的在前，再左大的在前
{
    return (x.r != y.r) ? (x.r<y.r) : (x.l>y.l);
}

int main()
{
    int n, rad;
    int d = 1;
    while (~scanf("%d%d", &n, &rad), (n&&rad))
    {
        int x, y;
        int f = 1;
        for (int i = 0; i<n; i++)
        {
            scanf("%d%d", &x, &y);
            if (rad >= y)//精彩的初始化，从二维问题变成一维问题
            {
                p[i].l = x - sqrt((double)rad*rad - (double)y*y);
                p[i].r = x + sqrt((double)rad*rad - (double)y*y);
            }
            else
            {
                f = 0;
            }
        }
        if (!f)
        {
            printf("Case %d: -1
", d++);
            continue;
        }
        sort(p, p + n, cmp);

        double End = p[0].r;
        int cnt = 1;
        for (int i = 1; i<n; i++)
        {
            if (End<p[i].l)
            {
                cnt++;
                End = p[i].r;
            }
        }
        printf("Case %d: %d
", d++, cnt);
    }
}