POJ3069 Saruman's Army Greedy（贪心）

Saruman's Army

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2527 Accepted: 1261

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

0 310 20 2010 770 30 1 7 15 20 50-1 -1

Sample Output

24

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.

题意：给你n个点，然后让你标记其中尽可能少的点，使得n个点都处于被标记点左右不超过R的区间内。

贪心：相当于n个球装进篮子，每个篮子最少只能装m个，如何把篮子利用率最高?那么把每个篮子都利用到最高装m个球，再最后一个篮子装下余下的球即可。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn =1005;int main(){    int n,r,x[maxn];    while(~scanf("%d%d",&r,&n)&&r>=0)    {        int ans=0;int i;        for(i=0;i<n;i++) scanf("%d",&x[i]);        sort(x,x+n);         i=0;        while(i<n)        {            int s=x[i++];            while(i<n&&x[i]<=s+r) i++;            int p=x[i-1];            while(i<n&&x[i]<=p+r) i++;            ans++;        }        printf("%d\n",ans);    }    return 0;}