poj 3069 Saruman's Army（贪心）

Saruman's Army

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 2

Problem Description

Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units, and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units of some palantir.

 

Input

<span lang="en-us"><p>The input test file will contain multiple cases. Each test case begins with a single line containing an integer <i>R</i>, the maximum effective range of all palantirs (where 0 ≤ <i>R</i> ≤ 1000), and an integer <i>n</i>, the number of troops in Saruman’s army (where 1 ≤ <i>n</i> ≤ 1000). The next line contains n integers, indicating the positions <i>x</i>₁, …, <i>x_n</i> of each troop (where 0 ≤ <i>x_i</i> ≤ 1000). The end-of-file is marked by a test case with <i>R</i> = <i>n</i> = −1.</p></span>

 

Output

<p>For each test case, print a single integer indicating the minimum number of palantirs needed.</p>

 

Sample Input

0 3 10 20 20 10 7 70 30 1 7 15 20 50 -1 -1

 

Sample Output

2 4

 

题目大意：数轴上有些点，每个点可以放个什么鬼东西，可以覆盖R范围，问最少需要多少个这个东西

题目分析：是挑战那本书上的题，从左往右扫，找圆心位置即可

 #include <iostream>
 #include<cstring>
 #include <string>
 #include <algorithm>
 using namespace std;
 bool cmp(int x, int y)
 {
     return x < y;
 }
 int main()
 {
     int r, n;
     int a[];
     while (cin >> r >> n)
     {
         if (r == - && n == -) break;
         int i;
         for (i = ; i <= n; i++)
         {
             cin >> a[i];
         }
         sort(a + , a + n + , cmp);
         int num = ;
         int x;
         i =;
         while (i <= n)
         {
             x = a[i++];
             while (i <= n && a[i] <= x + r) i++;//一直向有前进直到距s的距离大于r的点
             int p = a[i - ];//p是新加上标记的点的位置
             while (i <= n && a[i] <= p + r) i++;//一直向有前进直到距p的距离大于r的点
             num++;
         }
         cout << num << endl;
     }
     return ;
 }