【Poj】-3069-Saruman's Army（贪心）

最新推荐文章于 2020-06-14 23:21:47 发布

Bear1997

最新推荐文章于 2020-06-14 23:21:47 发布

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分类专栏：贪心算法思维

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探讨了如何利用最少数量的监视石（palantirs）来确保萨鲁曼军队中的每个成员都能处于监视石的有效范围内。通过算法确定最佳放置位置，实现全面覆盖。

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Saruman's Army

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7611		Accepted: 3888

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

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.

题意：n个数尽可能标记少的数使得每个数都在与标记的数相距为 r 的范围内。

题解：从左往右找数找到标记可以把第一个数包住且最远的数，把他标记，同理右边相距为 r 的数也是被包住的。再从没有被包住的第一个数开始找到距他最远且可以包住它的进行标记。

只想说书上的代码写的好巧，自己看了书的解释还是写不对，总之就是做题少，加油吧少年！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[1010];
int main()
{
	int d,n,num;						//d范围，n个数 
	while(~scanf("%d %d",&d,&n)&&(d!=-1&&n!=-1))
	{
		memset(a,0,sizeof(a));
		num=0;
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		sort(a+1,a+n+1);
		for(int i=1;i<=n;)
		{
			int now=a[i++];				//没有被覆盖的最左边的点的位置 
			while(i<=n&&a[i]<=now+d)
				i++; 				//一直向右找，找到据最左边点距离大于d的点 
			int p=a[i-1];				//加上标记的点是a[i-1] 
			while(i<=n&&a[i]<=p+d)		//同理被标记的点的右边也要处理，知道不可以被标记的点覆盖 
				i++;			//就是没有这一步(处理被标记点的右边)，怎么都不对！！！ 
			num++;				//记录被标记的次数 
		}
		printf("%d\n",num);
	}
	return 0;
}