POJ 3069 Saruman's Army(贪心)

 Saruman's Army

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

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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。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <string>

 5 #include <algorithm>

 6 #include <cctype>

 7 #include <cmath>

 8 #include <queue>

 9 #include <map>

10 #include <cstdlib>

11 using    namespace    std;

12 

13 int    main(void)

14 {

15     int    r,n;

16     int    s[1005];

17     int    ans,left,i,mid;

18 

19     while(scanf("%d%d",&r,&n) && (r != -1 && n != -1))

20     {

21         for(i = 0;i < n;i ++)

22             scanf("%d",&s[i]);

23         sort(s,s + n);

24 

25         ans = i = 0;

26         while(i < n)

27         {

28             left = s[i ++];

29             for(;i < n && s[i] - left <= r;i ++);

30             left = s[i - 1];

31             ans ++;

32             for(;i < n && s[i] - left <= r;i ++);

33         }

34         printf("%d\n",ans);

35     }

36 

37     return    0;

38 }