POJ 3176 Cow Bowling: Simple DP

Cow Bowling

http://poj.org/problem?id=3176

Time limit:1000ms

Memory limit:65536k

Description

The cows don ' t use actual bowling balls when they go bowling. They each take A/number (in the range 0..99), though, and line up in a standard bowling-pin-like as this:

          7



        3   8



      8   ,   1 0 2 7 4 4 4 5 2 6 5

Then the other cows traverse the triangle starting the ' its tip and moving ' down ' to one of the two diagonally adjacent Co WS until the "bottom" row is reached. The cow ' s score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins this frame. Given a triangle with n (1 <= n <=) rows, determine the highest possible sum achievable.

Input

Line 1: A single integer, N Lines 2..n+1:line i+1 contains i space-separated integers that represent row I of the Triangl E.

Output

Line 1:the largest sum achievable using the traversal rules

Sample Input

5
7
3 8
8 1
0 2 7 4 4 4 5 2-
6 5

Sample Output

30

Hint

Explanation of the sample:

          7

         *

        3   8

       *

      8   1   0

       *   2 7 4 4 * 4 5 2   6   5

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The highest score is achievable by traversing the cows as shown.

Source

Usaco December Bronze

Memory Search:

/*47ms,1356kb*/
    
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
using namespace Std;  
const int MAXN = 355;  
    
int A[MAXN][MAXN], dp[maxn][maxn], N;  
    
int f (int i, int j)  
{  
    if (Dp[i][j] >= 0) return dp[i][j];  
    if (i = = n-1) return dp[i][j] = a[i][j];  
    return dp[i][j] = A[i][j] + max (f (i + 1, j), F (i + 1, j + 1));  
}  
    
int main ()  
{  
    int i, J;  
    scanf ("%d", &n);  
    for (i = 0; i < n; ++i) for  
        (j = 0; J <= i; ++j)  
            scanf ("%d", &a[i][j]);  
    Memset (DP,-1, sizeof (DP));  
    printf ("%d", f (0, 0));  
    return 0;  
}

Recursive push:

/*16ms,868kb*/
    
#include <cstdio>  
#include <algorithm>  
using namespace std;  
const int MAXN = 355;  
    
int A[MAXN][MAXN];  
    
int main ()  
{  
    int n, I, J;  
    scanf ("%d", &n);  
    for (i = 0; i < n; ++i) for  
        (j = 0; J <= i; ++j)  
            scanf ("%d", &a[i][j]);  
    for (i = n-2 i >= 0;-i)  
        for (j = 0; J <= i; ++j)  
            a[i][j] + = max (a[i + 1][j), A[i + 1][j + 1]);  
    printf ("%d", a[0][0]);  
    return 0;  
}