POJ 3069 Saruman's Army

Saruman's Army

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5526		Accepted: 2829

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

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.

Source

Stanford Local 2006

 

题目大意：一个一维数组，一个半径为R的圆可以罩住几个点，给你半径和这些点，问用最少几个圆可以把所有点都罩住。

 

思路：模拟一个圆在滑动，开始圆数位。

3种情况，1：下一个点在现在这个圆内，且当圆心移动到下一个点时，最左边的点还在圆上，这时把圆心移向这个点。

2：下一个点在现在这个圆内，但是当圆心移向下一个点时，最左边的点已经出圆了，这时什么都不做继续看下一个点。

3：下一个点不在现在这个圆内了，这时启动一个新的圆，圆数加1。

 

#include<math.h>
#include<stdio.h>
#include<queue>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define N 1234

int main()
{
    int r,n,a[N];
    while(~scanf("%d%d",&r,&n))
    {
        if(r==-1&&n==-1)break;

        int d=2*r;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int c=a[0];
        int b=a[0];
        int cnt=1;
        for(int i=1;i<n;i++)
        {
            if( (a[i]-c)<=r && (a[i]-b<=r) )
            {
                c=a[i];
            }

            else if( (a[i]-c)<=r && (a[i]-b>r) )
            {

            }
            else if( (a[i]-c)>r && (a[i]-b>r) )
            {
                c=a[i];
                b=a[i];
                cnt++;
            }
        }
        cout<<cnt<<endl;

    }
    return 0;
}