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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4487 | Accepted: 2314 |
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, …, xn of each troop (where 0 ≤ xi ≤ 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
题意：一条直线上有n个点，现需要标记一些点，来满足每个点周围距离r以内必须有被标记的点（自己也算），问这些点至少要标记多少个才能满足要求。
解析：从左向右，一直贪心。从最左边的点开始，然后略过r以内所有点，然后再以距离r以内的最右边那个点标记，然后再将被标记的点一直向右贪心，重复上述操作即可。
AC代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int a[1002];
int main(){
#ifdef sxk
freopen("in.txt", "r", stdin);
#endif //sxk
int r, n;
while(scanf("%d%d", &r, &n)!=EOF && !(r==-1 && n==-1)){
for(int i=0; i<n; i++) scanf("%d", &a[i]);
sort(a, a+n); //从小到大排序
int i = 0, ans = 0;
while(i < n){
int s = a[i++];
while(i < n && a[i] <= s+r) i ++; //略过距离r以内的点
int p = a[i-1]; //标记r以内的最右边那个点
while(i < n && a[i] <= p+r) i ++; //一直向右贪心
ans ++;
}
printf("%d\n", ans);
}
return 0;
}
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原文地址：http://blog.csdn.net/u013446688/article/details/43017693
