Saruman ' s Army

Part I: Topics

Title Link: http://poj.org/problem?id=3069

Saruman ' s Army

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6839		Accepted: 3516

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.

Part II: Ideas

Greedy: 1, from small to large sort. Starting at the far left, starting with an not contained point and looking back at a point as the center, ensure that the circle with the radius of R must contain a starting point.

2, look for the nearest point with the center but at the outside of the circle as a new starting point, and then find the center.

3, repeat until the far right.

4, Note: 1, there may be duplicate points that can be sorted when going heavy. (This is used to "de-weigh" the distance as an integer when comparing distances).

2, if the last point is a new starting point, that is, to the center of their own.

Part III: Code

#include <iostream>#include<algorithm>using namespacestd;intMain () {intn,m; intarmy[1001];  while(cin>>m>>n,m!=-1&&n!=-1)//when both N and M are-1 o'clock ends    {         for(intI=0; i<n;i++) {cin>>Army[i]; } sort (Army,army+N);//sort from small to large        intCount=0; //starting from the far left, take a non-contained point as a starting point, and find as many points as possible//as the center of the circle. But the starting point must be included.        When you find the center, you need to find the next starting point. //This starting point must be outside the circle.         inttemp=army[0]; //Flag 1 To find the starting point        intflag=1;  for(intI=1; i<n;i++)        {            //when the starting point exists and the current point does not contain a starting point, then the previous point is the center of the Circle            if(flag&&army[i]-temp>0&&army[i]-temp>m) {count++; Temp=army[i-1];//Center                if(army[i]-temp>m)//The current point cannot be enclosed by a circle{Temp=army[i];//take this point as a new starting point                }                Else{flag=0;//no new starting point has been found since the center was found                    Continue; }            }            if(flag==0&&army[i]-temp>m) {temp=Army[i]; Flag=1; }        }        if(flag)//if flag is 1, it indicates a new starting point, and this is the last point. Give it a mark .{Count++; } cout<<count<<Endl; }    return 0;}

Saruman ' s Army