getting WA when I try to submit mkbudget spoj - c++

Question

I cannot understand/think of a case where my code fails to give correct output.
Link to the problem: http://www.spoj.pl/problems/MKBUDGET/
The problem clearly has a DP solution. I am posting my solution below:
#include <algorithm>
#include <cstdio>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<vector <int> > opt;
void compute_opt(vector<int> A,int n,int hire,int fire,int sal,int max_a)
{
for(int i = A[0]; i <= max_a; i++) //for num workers in 1st month
opt[0][i] = i*(hire + sal);
for(int i = 1; i < n; i++) //num of months
for(int j = A[i]; j <= max_a; j++) //num of workers for ith month >=A[i] and <= max_a
{
opt[i][j] = opt[i-1][A[i-1]] + j*sal + (A[i] > A[i-1] ? (A[i]-A[i-1])*hire : (A[i-1] - A[i])*fire);
for(int k = A[i-1]; k <= max_a; k++)
opt[i][j] = min(opt[i][j], opt[i-1][k] + j*sal + (j>k ? (j-k)*hire : (k-j)*fire));
}
}
int ans(vector<int> A, int n, int max_a)
{
int ret = opt[n-1][A[n-1]];
for(int i = A[n-1]; i <= max_a; i++)
ret = min (ret, opt[n-1][i]);
return ret;
}
int main()
{
vector<int> A;
int n, hire, fire, sal,max_a, c = 1;
while(1)
{
cin >> n;
if(n == 0)
break;
A.clear();
opt.clear();
max_a = 0;
cin >> hire >> sal >> fire;
A.resize(n);
for(int i = 0; i < n; i++)
{cin >> A[i];
max_a = max(max_a,A[i]);
}
opt.resize(n);
for(int i = 0; i < n; i++)
opt[i].resize(max_a + 2);
compute_opt(A,n,hire,fire,sal,max_a);
cout << "Case " << c << ", cost = $" << ans(A,n,max_a) << endl;
c++;
}
return 0;
}
I am getting correct answers for the two sample test cases but I get a WA when I submit. Any help ?

Answer 1

OK, so your problem is that you disallow the case where you hire any number of employees between A[i] and A[i - 1]. Maybe it's a good idea to fire some unneeded employees, but not all. That's why you get WA. I modified your code and got it accepted:
void compute_opt(vector<int> A,int n,int hire,int fire,int sal,int max_a)
{
// Fill all disallowed entries with infinity
for (int i = 0; i < A[0]; ++i)
opt[0][i] = 1000000000;
for(int i = A[0]; i <= max_a; i++) //for num workers in 1st month
opt[0][i] = i*(hire + sal);
for(int i = 1; i < n; i++)
for(int j = 0; j <= max_a; j++)
{
// No need for special case handling,
//just check all previous numbers of employees
opt[i][j] = 1000000000;
if (A[i] > j) continue;
for(int k = 0; k <= max_a; k++)
opt[i][j] = min(opt[i][j],
opt[i-1][k] + j*sal + (j>k ? (j-k)*hire : (k-j)*fire));
}
}
By the way, there's a "greedier" solution than the one you have that does not depend on the number of employees being small (so that the table can be allocated).