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[1005];
    int    ans,left,i,mid;

    while(scanf("%d%d",&r,&n) && (r != -1 && n != -1))
    {
        for(i = 0;i < n;i ++)
            scanf("%d",&s[i]);
        sort(s,s + n);

        ans = i = 0;
        while(i < n)
        {
            left = s[i ++];
            for(;i < n && s[i] - left <= r;i ++);
            left = s[i - 1];
            ans ++;
            for(;i < n && s[i] - left <= r;i ++);
        }
        printf("%d\n",ans);
    }

    return    0;
}