POJ 3069 Saruman's Army （模拟）

题目连接

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.

分析：
我们从最左边开始考虑。对于这个点，到距其R以内的区域内必须要有带有标记的点。（此点位于最左边，所以显然）带有标记的这个点一定在此点右侧（包含这个点自身）。

于是，究竟要给哪个点加上标记呢？答案应该是从最左边的点开始，距离为R以内的最远的点，因为更左的区域没有覆盖的意义，所以应该尽可能覆盖更靠右的点。

加上了第一个标记后，剩下的部分也用同样的办法处理。对于添加了符号的点右侧相距超过R的下一个点，采用同样的方法找到其右侧R距离以内最远的点添加标记。在所有的点都被覆盖之前不断重复这一过程。

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[1009];
int r,n;
void solve()
{
    int i=0,ans=0;
    while(i<n)
    {
        int s=a[i++];
        while(i<n&&a[i]<=s+r) i++;///一直找到那个与当前这个点的距离超过r的点
        int p=a[i-1];///确定下一个标记点
        while(i<n&&a[i]<=p+r) i++;///再找到距离标记点距离为r的点
        ans++;
    }
    printf("%d\n",ans);
}
int main()
{
    while(~scanf("%d%d",&r,&n))
    {
        if(r==-1&&n==-1)///循环结束条件
            break;
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        solve();
    }
    return 0;
}