POJ1328

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

解题思路：

贪心算法，从左到右建立雷达，要尽量多地覆盖岛屿（对某一个岛屿而言，雷达建在能覆盖该岛屿的最右侧位置上）。

以岛屿为圆心，以d为半径画圆，如果画出的圆与x轴没有交点，则不能实现。存在交点的话，计算出第i个岛屿与x轴的交点坐标保存在结构体数组rad[i].sta与rad[i].end中。

对结构体数组排序，按照rad[i].end（因为确定雷达位置时依照end的值，保证将验证的未访问的点的end在当前end的右侧，若sta在当前sta的左侧，满足覆盖条件）进行升序排列。

然后一次从左到右找雷达。对于rad[i].end为当前最右边的左坐标，对于下一个岛屿，如果rad[j].sta<ran[i].end，则不需要建立新的雷达，需要更新下一个岛屿。否则需要建立新的雷达。

cpp源代码：

#include<iostream>#include<math.h>#include <stdio.h>#include<algorithm>#include<memory.h>using namespace std;const int Max = 1005;struct{    int x, y;}isl[Max];    //  小岛的数据。struct data{    float sta, end;}rad[Max];    //  小岛所对应雷达的数据。bool cmp(data a, data b){    if(a.end < b.end)        return true;    else        return false;}int main(){    int n, d, t = 1;    while(cin >> n >> d && n != 0)    {        int i, j, max_y = 0;        for(i = 0; i < n; i ++)        {            cin >> isl[i].x >> isl[i].y;            if(isl[i].y > max_y)                max_y = isl[i].y;        }        getchar();        getchar();  //  PE了两次。        cout << "Case " << t ++ << ": ";        if(max_y > d || d < 0)        {            cout << -1 << endl;            continue;        }        float len;        for(i = 0; i < n; i ++)        {   //  求出小岛所对应雷达的可能覆盖范围。            len = sqrt(1.0 * d * d - isl[i].y * isl[i].y);            rad[i].sta = isl[i].x - len;            rad[i].end = isl[i].x + len;        }        sort(rad, rad + n, cmp);   //  根据rad的end值进行排序。        int ans = 0;        bool vis[Max];        memset(vis, false, sizeof(vis));        for(i = 0; i < n; i ++)        {   //  类似的活动选择。            if(!vis[i])            {                vis[i] = true;                for(j = 0; j < n; j ++)                    if(!vis[j] && rad[j].sta <= rad[i].end)                        vis[j] = true;                    ans ++;            }        }        cout << ans << endl;    }    return 0;}