POJ 3069 Saruman's Army (簡單貪心)

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Saruman‘s Army
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5343   Accepted: 2733

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 integerR, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integern, 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 withR = 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.

Source

Stanford Local 2006

題目連結:http://poj.org/problem?id=3069

題目大意:數軸上有些點,每個點可以放個什麼鬼東西,可以覆蓋R範圍,問最少需要多少個這個東西

題目分析:好像是挑戰那本書上的題,從左往右掃,找圓心位置即可



#include <cstdio>#include <iostream>#include <algorithm>using namespace std;int x[1005];int main(){int r,n;while(scanf("%d %d", &r, &n) != EOF && (r != -1 && n != -1)){for(int i = 0; i < n; i++)scanf("%d", &x[i]);sort(x, x + n);int ans = 0, i = 0;while(i < n){int a1 = x[i++];while(i < n && x[i] <= a1 + r)i ++;int a2 = x[i - 1];while(i < n && x[i] <= a2 + r)i++;ans++;}printf("%d\n", ans);}}



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POJ 3069 Saruman's Army (簡單貪心)

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