POJ 3069 Saruman's Army(贪心)

 Saruman's Army

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3069

Appoint description: 
System Crawler  (2015-04-27)

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

The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x₁, …, x_n of each troop (where 0 ≤ x_i ≤ 1000). The end-of-file is marked by a test case with R = n = −1.

Output

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

Sample Input

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

Sample Output

2
4

尽量使每个球覆盖的位置最大。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <cctype>
 #include <cmath>
 #include <queue>
 #include <map>
 #include <cstdlib>
 using    namespace    std;

 int    main(void)
 {
     int    r,n;
     int    s[];
     int    ans,left,i,mid;

     while(scanf("%d%d",&r,&n) && (r != - && n != -))
     {
         for(i = ;i < n;i ++)
             scanf("%d",&s[i]);
         sort(s,s + n);

         ans = i = ;
         while(i < n)
         {
             left = s[i ++];
             for(;i < n && s[i] - left <= r;i ++);
             left = s[i - ];
             ans ++;
             for(;i < n && s[i] - left <= r;i ++);
         }
         printf("%d\n",ans);
     }

     return    ;
 }