poj 3069 Saruman's Army（贪心）

Saruman's Army

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 2

Problem 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

<span lang="en-us"><p>The input test file will contain multiple cases. Each test case begins with a single line containing an integer <i>R</i>, the maximum effective range of all palantirs (where 0 ≤ <i>R</i> ≤ 1000), and an integer <i>n</i>, the number of troops in Saruman’s army (where 1 ≤ <i>n</i> ≤ 1000). The next line contains n integers, indicating the positions <i>x</i>₁, …, <i>x_n</i> of each troop (where 0 ≤ <i>x_i</i> ≤ 1000). The end-of-file is marked by a test case with <i>R</i> = <i>n</i> = −1.</p></span>

 

Output

<p>For each test case, print a single integer indicating the minimum number of palantirs needed.</p>

 

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：数轴上有些点，每个点可以放个什么鬼东西，可以覆盖R范围，问最少需要多少个这个东西

题目分析：是挑战那本书上的题，从左往右扫，找圆心位置即可

#include <iostream>
#include<cstring>
#include <string>
#include <algorithm>
using namespace std;
bool cmp(int x, int y)
{
    return x < y;
}
int main()
{
    int r, n;
    int a[1005];
    while (cin >> r >> n)
    {
        if (r == -1 && n == -1) break;
        int i;
        for (i = 1; i <= n; i++)
        {
            cin >> a[i];
        }
        sort(a + 1, a + n + 1, cmp);
        int num = 0;
        int x;
        i =1;
        while (i <= n)
        {
            x = a[i++];
            while (i <= n && a[i] <= x + r) i++;//一直向有前进直到距s的距离大于r的点
            int p = a[i - 1];//p是新加上标记的点的位置
            while (i <= n && a[i] <= p + r) i++;//一直向有前进直到距p的距离大于r的点
            num++;
        }
        cout << num << endl;
    }
    return 0;
}