POJ 3069 贪心

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

求每段区间的中间位置，先扫左边，从第一个军队位置开始，扫到n区间内的最后一个军队的位置，把灯塔放这，再扫右边区间的最后一个位置。由于是对称结构，一次扫一对区间才能求出来，并不是每个区间独立，所以在while里面要每次都判断左右两区间，这就简单很多了。像我这种新手，刚开始白痴的想着，每次只判断一次区间。根本不可能做出来。要学会用纸画图解题，看题是没用的的。

#include <string>#include <cstring>#include <iostream>#include <algorithm>#include <stack>#include <cstdio>using namespace std;int a[1010];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF&&n!=-1&&m!=-1)    {        int i;        for(i=0;i<m;i++)        {            scanf("%d",&a[i]);        }        sort(a,a+m);        i=0;        int sum=0;        while(i<m)        {            int s=a[i++];            while(i<m&&a[i]<=s+n) i++;            int p=a[i-1];            while(i<m&&a[i]<=p+n) i++;            sum++;        }        printf("%d\n", sum);    }}