poj 3069 Saruman's Army

Posted 2020-11-05 给杰瑞一块奶酪~

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13285		Accepted: 6659

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

给出r和n，r是palantir能覆盖的范围，n是troop个数，要求用最少的palantir,使得每个军队都会在掌握之中。

贪心思想，首先按照位置对n个数从小到大排序，palantir按在的位置j应当保证他左边有尽可能多的尚且没有被前一个palantir覆盖的troop，都能被他覆盖，然后右边能被他覆盖的troop统统划分给他。循环这样的操作。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define MAX 1002
using namespace std;
int s[MAX],n,r;
int main()
{
    while(~scanf("%d%d",&r,&n) && r != -1 && n != -1)
    {
        for(int i = 0;i < n;i ++)
        {
            scanf("%d",&s[i]);
        }
        sort(s,s + n);
        int c = 0,i = 0;
        while(i < n)
        {
            c ++;
            int j = i;
            while(j + 1 < n && s[j + 1] - s[i] <= r)j ++;
            i = j;
            while(j + 1 < n && s[j + 1] - s[i] <= r)j ++;
            i = j + 1;
        }
        printf("%d\n",c);
    }
}

View Code