B - B Saruman's Army(贪心)

在一条直线上，有n个点。从这n个点中选择若干个，给他们加上标记。对于每一个点，其距离为R以内的区域里必须有一个被标记的点。问至少要有多少点被加上标记

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

Hint

In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.

In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.

题解：我们从最左边的开始考虑。对于这个点，到距其R以内的区域必须要有带有标记的点。带有标记的点一定在其右侧（包含这个点本身）。给从最左边开始，距离为R以内的最远的点加上标记，尽可能的覆盖更靠右边的点。对于添加了标记的点右侧相距超过R的下一个点，采用同样的方法找到最右侧R距离以内最远的点添加标记。在所有点都被覆盖之前不断重复这一过程。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[1010];
int main()
{
    int n,r;
    while(~scanf("%d%d",&r,&n)&&(n+r)!=-2)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        scanf("%d",&dp[i]);
        int ans=0;
        sort(dp+1,dp+n+1);
        int i=1;
        while(i<=n)
        {
            int s=dp[i++];
            while(i<=n&&dp[i]<=s+r)
                i++;
            int p=dp[i-1];
            while(i<=n&&dp[i]<=p+r)
                i++;
            ans++;
        }
        printf("%d\n",ans);
    }
}