Poj 3069 Saruman's Army【贪心】

Saruman's Army

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7931 Accepted: 4054

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：

开始输入两个变量，r，n，表示每个点可以覆盖的区域的半径（可以覆盖其左边和右边）【选择了之后才可以覆盖】，以及点的个数。问最少选择多少个点就能够使得所有点被覆盖进去。

思路：

1、首先将输入的点去重，然后从小到大排序。

2、然后对应排序结束之后的点，从第一个点开始算起设定为l，从这个点一直向右寻找到第一个与这个点距离大于r的点j，然后设定mid为j-1，然后再从mid这个点开始向右寻找到第一个与这个点距离大于r的点jj，然后从第一个点到第jj个点就被mid这个点所覆盖了。然后这时候再设定l=jj，继续如此一直进行，直到不能进行为止，维护其最优解即可。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;int vis[15000];int a[150000];int main(){    int r,n,tmpn;    while(~scanf("%d%d",&r,&tmpn))    {        if(r==-1&&tmpn==-1)break;        n=0;        memset(vis,0,sizeof(vis));        for(int i=1;i<=tmpn;i++)        {            int tmp;            scanf("%d",&tmp);            if(vis[tmp]==1)continue;            a[n++]=tmp;            vis[tmp]=1;        }        sort(a,a+n);        int output=0;        int l=0;        while(l<n)        {            int mid=-1;            int rr=-1;            for(int j=l+1;j<n;j++)            {                if(a[j]-a[l]>r)                {                    mid=j-1;break;                }            }            if(mid==-1)            {                output++;break;            }            if(mid==l)            {                output++;l++;                continue;            }            for(int j=mid+1;j<n;j++)            {                if(a[j]-a[mid]>r)                {                    rr=j;break;                }            }            if(rr==-1)            {                output++;break;            }            if(rr==mid)            {                output++;l=mid+1;                continue;            }            l=rr;            output++;        }       printf("%d\n",output);    }}