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[codejam] ant stack

admin 11月 13, 2020 0

Link to the problem

I think this is a good dynamic programming problem for exercise🤔.
One tricky point is that we can find the maximum number of ants that we can achieve when weight $ leq 10^{9} $ is around 138, which is much smaller than $10^{5}$.
dp[i][j] means the minimum weight when we use ants from 1 to i to get a pile containing j ants. And the transition function is like
dp[i][j] = min( dp[i-1][j], dp[i-1][j-1] + weights[i] (if 6*weights[i] $geq$ dp[i-1][j-1]))

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#include <algorithm>
#include <set>
#include <math.h>
#include <string>
#include <set>
#include <map>
#include <cstring>
using namespace std;
#define ll  long long


const int maxn = 100005;
const int maxant = 150;
ll weights[maxn];
ll dp[maxn][maxant+5];

int ()
{
	int T;
	scanf("%d",&T);
	int N;
	for(int t = 1; t <= T;t++)
	{
		printf("Case #%d: ",t);
		scanf("%d",&N);

		for(int i = 0 ;i <= N;i++)
			for(int j = 0; j <= maxant;j++)
			{
				dp[i][j] = 0x3f3f3f3f3f3f3f3f;
			}

		for(int i = 1;i <= N;i++)
		{
			scanf("%lld",&weights[i]);
		}
		for(int i = 0; i<= N;i++)
		{
			dp[i][0] =0;
		}
		for(int i = 1;i <= N;i++)
		{
			for(int j = 1;j <= maxant;j++)
			{
				if(dp[i][j] > dp[i-1][j])
					{
						dp[i][j] = dp[i-1][j];
					}
				if(6*weights[i] >= dp[i-1][j-1])
				{
					if(dp[i][j] > dp[i-1][j-1] + weights[i])
					dp[i][j] = dp[i-1][j-1] + weights[i];

				}
			}
		}
		for(int i = maxant;i >= 1;i--)
		{
			if(dp[N][i] != 0x3f3f3f3f3f3f3f3f)
			{
				printf("%dn",i);
				break;
			}

		}

	}

return 0;
}
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