ACM: uva 10382

Problem E Watering Grass

n sprinklers are installed in a horizontal strip of grass l meters long and w meters wide. Each sprinkler is installed at the horizontal center line of the strip. For each sprinkler we are given its position as the distance from the left end of the center line and its radius of operation.

What is the minimum number of sprinklers to turn on in order to water the entire strip of grass?

Input

Input consists of a number of cases. The first line for each case contains integer numbers n, l and w with n <= 10000. The next n lines contain two integers giving the position of a sprinkler and its radius of operation. (The picture above illustrates the first case from the sample input.)

 

Output

For each test case output the minimum number of sprinklers needed to water the entire strip of grass. If it is impossible to water the entire strip output -1.

Sample input

8 20 2

5 3

4 1

1 2

7 2

10 2

13 3

16 2

19 4

3 10 1

3 5

9 3

6 1

3 10 1

5 3

1 1

9 1

 

Sample Output

6

2

-1

题意: 在一个l*w的矩形区域内, 现在又n个半径不一的圆形并且圆心都在矩形区域水平中线上, 现在

      要你从中选出尽可能的少的圆形覆盖整个矩形区域.

解题思路:

      1. 值考虑半径大于2/w的圆形, 并且将圆形左右覆盖的区间计算出来. 如下图:

      2. 结果上面的装换: 圆形区域变成了[pi-t, pi+t]区间了, 问题变成最少区间覆盖问题.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define MAX 10005

struct node
{
 double left, right;
 bool operator <(const node &a) const
 {
  return left < a.left;
 }
}p[MAX];

int n, num;
double l, w;

int main()
{
 int i;
// freopen("input.txt", "r", stdin);
 while(scanf("%d %lf %lf", &n, &l, &w) != EOF)
 {
  num = 0;
  double po, r;
  for(i = 0; i < n; ++i)
  {
   scanf("%lf %lf", &po, &r);
   if(r <= w/2) continue;
   double t = sqrt(r*r-w*w/4.0);
   p[num].left = po-t;
   p[num++].right = po+t;
  }

  sort(p, p+num);
  double left = 0, right = 0;
  bool flag = false;
  int result = 0;
  i = 0;
  if(p[0].left <= left)
  {
   while(i < num)
   {
    int j = i;
    while(j < num && left >= p[j].left)
    {
     if(p[j].right > right)
      right = p[j].right;
     ++j;
    }
    if(j == i) break;
    result++;
    left = right;
    i = j;
    if(left >= l)
    {
     flag = true;
     break;
    }
   }
  }

  printf("%d\n", flag ? result : -1);
 }
 return 0;
}