POJ 3069 Saruman's Army

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最新推荐文章于 2023-07-25 08:25:19 发布

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解决Saruman'sArmy问题，通过贪心算法确定最少标记点数，确保每个点在其有效范围内至少有一个标记点。

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一道足足做了三天（当然其中还穿插着别的半成品555）的题。

Saruman's Army

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10006		Accepted: 5034

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

题目大意：在一个数轴上有N个点，现要在这N个点其中挑选一些点加上标记，要求所有有标记的点的R范围以内能包含所有点（换言之，对于每一个点，在它的R范围之内必须有一个被标记的点），求最少需要标记多少点。

思路：一上来感觉也看不出是个什么算法，但是画个图（高手请忽略此步骤）可以发现，有一些只有傻子才会做的操作：标记最左面或者最右面的点。不难看出这样会浪费距离，而且再一想每一个点都必须被包含所以完全可以线性顺次标记，所以很自然的想到贪心，代码附上。

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<ctime>

using namespace std;

const int MAXN=1010;

void work(int r,int n)
{
	int x[MAXN];
	int i;
	
	for (i=0;i<n;i++)
	{
		scanf("%d",&x[i]);
	}
	sort(x,x+n);
	int ans=0;
	i=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++;
	}
	printf("%d\n",ans);
}

int main()
{
	int a,b;
	scanf("%d%d",&a,&b);
	while (a!=-1 && b!=-1)
	{
		work(a,b);
		scanf("%d%d",&a,&b);
	}
    return 0;
}

这次数组下标从0开始！此处s是没有被覆盖的最左边的点的位置，p是每一次新加的点的位置。注意while循环中i<n，没有等号！！博主在这个地方提供了一次分母。。。

要加速了不然没法每天一题了233。