poj-3065

                                                                                                                    Saruman's Army
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5711 Accepted: 2927
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

题目来源:http://poj.org/problem?id=3069

考察点:贪心算法

题目大意:给你n个数,从这n个数中选取最少的数,做上标记,使对每一个数,在它距离为r的范围内都有被标记的数.求最少选取几个数.

解题思路:将n个数从小到大排序,然后从最左边的点开始(记为a点),寻找一个距a点r范围内最远的点,那么该点就是因该标记的点,记为b,因为它的覆盖面最大,满足最优情况,剩下的点也这样处理.寻找下一个点时,跳过(b-r,b+r).

代码如下:

#include<stdio.h>#include<algorithm>using namespace std;int x[1010];int ans;int r,n;int main(){    int i,j,y;while(scanf("%d%d",&r,&n)){if(n==-1&&r==-1)break;for(i=0;i<n;i++)scanf("%d",&x[i]);sort(x,x+n);ans=0;int t,p;int flag=0;for(i=0;i<n;){   for(j=i+1;j<n;j++)   {   if(x[i]+r<x[j])break;       }       for(p=j;p<n;p++)       {       if(x[j-1]+r<x[p])  break;       }       ans++;  i=p;   }printf("%d\n",ans);}return 0;}