POJ 1328

Radar Installation

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 90690 Accepted: 20370
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
个人理解：
转换为区间，降维，进行贪心，当然越建左边对后面影响越好。排序是按照右端点为第一优先级，左端点为第二优先级。WA了无数次。。。不知道为什么copy了一份别人的代码。思路差不多，只不过没有按照左端点排序，加了一个标记，重复扫描区间。

#include<iostream>  #include<math.h>  #include<algorithm>  using namespace std;  const int Max = 1005;  struct  {      int x, y;  }isl[Max];   struct data  {      float sta, end;  }rad[Max];     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();        cout << "Case " << t ++ << ": ";          if(max_y > d || d < 0)          {              cout << -1 << endl;              continue;          }          float len;          for(i = 0; i < n; i ++)          {              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);           int ans = 0;          bool vis[Max];          memset(vis, false, sizeof(vis));          for(i = 0; i < n; i ++)          {                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;  }