【POJ3069】Saruman's Army

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
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.

Source
Stanford Local 2006

初探贪心类题目，感觉基本在考察智商思维能力
入手点是位于最边缘的士兵的位置（所以得先排个序），因为无论如何最边缘的士兵一定得在palantir的影响范围R内，可由此确定第一个palantir的位置
再思考一会就发现第一个palantir刚好影响不到的那个士兵的位置，和“最边缘的士兵的位置”是同一性质的东西，故可以写出循环

//Matsuri 
#include
#include
#define MAXN 2001
using namespace std;
int n,r,x[MAXN];int ans=0;
void solve()
{
	int i=1; //下标指示量，即下标现在指着哪一位士兵，初始时指着从左边开始第一位士兵  
	sort(x+1,x+n+1);
	while(i<=n)
	{
		int s=x[i]; //s为未被上一个r范围覆盖到的最左边的点
		//由于s必定要在下一个r范围内，故可根据s找下一个palantir
		while(i<=n && x[i]<=s+r) i++;
		int p=x[i-1]; //p为扛着palantir的士兵的位置，此时已经找到 
		while(i<=n && x[i]<=p+r) i++; 
		ans++;
	}	
	printf("%d\n",ans);
}
int main()
{
	while(scanf("%d%d",&r,&n)!=EOF)
	{
		if (r==-1 || n==-1) break;
		ans=0;
		for (int i=1;i<=n;i++) scanf("%d",&x[i]);
		solve();
	}
	return 0;
}