POJ3069 Saruman's Army

72 阅读2分钟

题目:

| Saruman's Army| Time Limit: 1000MS |   | Memory Limit: 65536K | | --------------------------- | - | ------------------------ | | Total Submissions: 8359 |   | Accepted: 4257 |DescriptionSaruman 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.InputThe 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 positionsx1, …, xn of each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case with R = n = −1.OutputFor each test case, print a single integer indicating the minimum number of palantirs needed.Sample Input0 3 10 20 20 10 7 70 30 1 7 15 20 50 -1 -1Sample Output``` 2 4

| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | - | - |

\[[Submit](http://poj.org/submit?problem_id=3069)]   \[Go Back]   \[[Status](http://poj.org/problemstatus?problem_id=3069)]   \[[Discuss](http://poj.org/bbs?problem_id=3069)]

代码:

```cpp
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int main()
{
    while(1)
    {
        int n,r,a[1010];
        scanf("%d %d",&r,&n);
        if(r==-1&&n==-1)return 0;
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int i=0,ans=0;
        while(i<n)
        {
            int s=a[i++];//先让s指向第一个点,s是没有被覆盖的最左的点的位置
            while(i<n&&a[i]<=s+r)i++;//一直向右前进知道距离s的距离大于r的点
            int p=a[i-1];//新加上的标记的点的位置
            while(i<n&&a[i]<=p+r)i++;//一直向右前进知道距离P的距离大于r的点
            ans++;//记录标记的点的个数
        }
        printf("%d\n",ans);
    }
    return 0;
}


PS:还是贪心的题,给了很多点给定一个半径,让选出最小的点的数量覆盖完这半径范围内所有点。