Poj-3069-saruman ' s Army (Java simple greedy)

Saruman ' s Army Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4699 Accepted: 2430 Description Saruman The white must leads 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 is carried by some troop in the army (i.e., Pala Ntirs is 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 Each of the Minions is within R units of some palantir. Input The input test file would contain multiple cases. Each test case begins with a containing a integer R, the maximum effective range of all Palantirs (w Here 0≤ R ≤1000), and a integer n, the number of troops in Saruman ' s Army (where 1≤ n ≤100 0). The next line contains n integers, indicating the positions x1, ..., xn 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 0 310 20 2010 770 30 1 7 15 20 50-1-1 Sample Output 24 Hint In the first test case, Saruman is a palantir at positions and 20. Here, note this 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), Position (Coverin g positions and), position, and position 70. Here, note this palantirs must be distributed among troops and is not allowed to ' free float. ' Thus, Saruman cannot place a palantir at position-cover the troops at positions and 70. Source Stanford Local 2006 first explain the topic (Bo Master English slag!) (Translated from the Challenge Program Design Competition): There are n dots on the line. The position of point I is XI. Select several of these n points and mark them with a tag. For each point, the area within the distance of R must have a labeled point (its own labeled point, which can be thought to have a marked point with its distance of 0). In cases where this condition is met, it is desirable to add markers for as few points as possible. How many points should be added to the mark at least? Restrictions:1.1<=n<=10002.  0<=r<=10003. 0<=xi<=1000 Examples: Input: r=10 n=6x={1,7,15,20,30,50} Output:3 /* * Starting from the leftmost point, for this point, there must be a marked point within the range of R. * This point is located on the far left, and obviously marked points must be right at this point! So from the leftmost point, the distance * is the farthest point within the R, for the point where the symbol is added to the right of the next point above R, you can think of this point as the first leftmost point, because these two points are equivalent, that is, the same method can be used! */import java.io.*;import java.util.*;p ublic class main{public static void Main (string[] args) {//TODO auto-generated met Hod Stubscanner input = new Scanner (system.in), while (Input.hasnext ()) {int R = Input.nextint (); int N = Input.nextint (); int x[] = new Int[n];for (int i = 0; i < N; i++) {X[i] = Input.nextint ();} Arrays.sort (X); int i = 0, ans = 0;while (i < n) {int s = x[i++];//always to the right until the distance from S is greater than the point of R (I < n && X[i] < = s + R) i++;//p is the position of the new tagged point int p = x[i-1];//goes right forward until the distance from P is greater than the point of R (i < N && X[i] <= p + r) i++;ans++;} System.out.println (ANS);}}

Poj-3069-saruman ' s Army (Java simple greedy)