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


Solution

题目大意:直线上有N个点,对应的点为Xi,对于任何一个点,以R为半径,它周围必须有被标记的点,在满足这个条件的情况下,求最少需要标记多少个点。
解题思路:
先对坐标进行升序排序。
1. 以a[0]为左端点,找满足a[i]>a[0]+r的第一个a[i],找到后以a[i-1]画圆。
2. 然后找到满足a[i-1]+r
3. 重复上述步骤,直到j=n。


Code

#include 
#include 
#include 
#include 
using namespace std;
#define N 1010
int a[N];

int main( )
{
    int r, n,j=1;
    memset(a, -1, sizeof(a));
    while (scanf("%d%d", &r, &n) && r != -1 && n != -1)
    {
        for (int i = 0; i scanf("%d", &a[i]);
        sort(a, a + n);
        int ans = 0;
        int j = 0;
        while (jint temp = a[j];
            while (a[j] - temp 1];
            while (a[j] - temp printf("%d\n", ans);
    }
    return 0;
}

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转载文章请注明:POJ 3069 – Saruman’s Army(贪心) - https://blog.i-ll.cc/archives/58

分类: Coding

Vectors

Vectors

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