Poj3069 Saruman's Army

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字数统计: 493 | 阅读时长 ≈ 2

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, …, xn of each troop (where 0 ≤ xi ≤ 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

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2
3
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5
0 3
10 20 20
10 7
70 30 1 7 15 20 50
-1 -1

Sample Output

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2
2
4

Analysis

一条直线上有n个点，从n各点中选若干个加上标记，对于每一个点，距离其R以内的区域里必须有一个被标记的点，问至少有多少点需要加上标记。

根据贪心的思想，使用以下步骤：

  1. 从最左点开始，这个点右侧R距离以内的区域必须有一个被标记的点，那么就给距离其R以内的离它最远的点加上标记。
  2. 找到这个被标记点右侧R距离外最近的一个点。
  3. 将此点作为最左点，重复1、2，直到所有点都被覆盖。

Code

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#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
	int r, n;
	int x[10000];
	while (cin >> r >> n)
	{
		if (r == -1 && n == -1)
			break;
		memset(x, 0, sizeof(x));
		for (int i = 0; i < n; i++)
		{
			cin >> x[i];
		}
		sort(x, x + n);
		int i = 0, ans = 0;
		while (i < n)
		{
			int s = x[i++];
			while (i < n&&x[i] <= s + r)
				i++;
			int p = x[i - 1];
			while (i < n&&x[i] <= p + r)
				i++;
			ans++;
		}
		cout << ans << endl;
	}
	return 0;
}