Saruman ' s Army (greedy)

Saruman ' s ArmyTime limit:MS Memory Limit:65536KB 64bit IO Form At:%i64d &%i64u Submit Status Practice POJ 3069

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 have a maximum effective range of R units, and must be carried by

Some troop in the army (i.e., Palantirs 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 his minions is within R units of some palantir.

Input

The input test file would 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 ≤100 0).

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

0 310 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 (covering positions), 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.

Code:

 #include <iostream> #include <cstring> #include <string> #include < Algorithm> #include <cstdio>using namespace std;struct fun{int l;//interval left int r;//interval right int x;//input point}a[1100];int an S;int N, r;bool cmp (struct fun A, struct fun b)//Custom comparison function, ordered from small to large on the left of the interval {return A.L<B.L;} int main () {while (~SCANF ("%d%d", &r, &n) && r! =-1 && n! =-1) {for (int i = 0; i < N        i++) {scanf ("%d", &a[i].x);            } for (int i = 0; i < n; i++) {a[i].l = A[i].x-r;        A[I].R = A[i].x+r;        } sort (A, a + N, CMP);        int L, R; ans = 1;        L = a[0].x, r = A[0].R;            for (int i = 1;i < n; i++) {if (A[I].L <= l) r = A[I].R;                else if (R < a[i].x) {ans++;                L = a[i].x;            R = A[I].R;    }} cout<< ans <<endl; } return 0;} 

Code Listing 2:

The following code copies a Daniel's.

/**  mainly examines the greedy algorithm, first, the data is sorted in ascending order, then, using the greedy algorithm to solve this problem, get the final result. */#include <iostream> #include <algorithm>using namespace std;int solve (int a[], int n,int r) {    int ans = 0, i = 0;    Sort (a,a+n);       The array is ascending    while (i < n)    {        int s = a[i++];     The point        while (I < n&&a[i] <= S + R) is i++ to the right until the distance from S is greater than R;        int k = a[i-1];     The point        while (I < n&&a[i] <= K + R) is i++ to the right until the distance from K is greater than R;        ans++;    }    return ans;} int main (void) {    int n,m;    while (cin>> n >> m,~n&&~m)    {        int a[m];        for (int i = 0; I! = m; ++i)        {            cin >> a[i];        }        cout << Solve (a,m,n) << Endl;    }    return 0;}

Saruman ' s Army (greedy)