【lightoj 1004 Monkey Banana Problem (简单动规，就是数字三角形嘛。。。不过发现有个小问题)】

1004 - Monkey Banana Problem

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You are in the world of mathematics to solve the great "Monkey Banana Problem". It states that, a monkey enters into a diamond shaped two dimensional array and can jump in any of the adjacent cells down from its current position (see figure). While moving from one cell to another, the monkey eats all the bananas kept in that cell. The monkey enters into the array from the upper part and goes out through the lower part. Find the maximum number of bananas the monkey can eat.

 

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Every case starts with an integer N (1 ≤ N ≤ 100). It denotes that, there will be 2*N - 1 rows. The i^th (1 ≤ i ≤ N) line of next N lines contains exactly inumbers. Then there will be N - 1 lines. The j^th (1 ≤ j < N) line contains N - j integers. Each number is greater than zero and less than 2¹⁵.

Output

For each case, print the case number and maximum number of bananas eaten by the monkey.

Sample Input	Output for Sample Input
2 4 7 6 4 2 5 10 9 8 12 2 2 12 7 8 2 10 2 1 2 3 1	Case 1: 63 Case 2: 5

Note

Dataset is huge, use faster I/O methods.

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SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)

 

 

 

 

// Project name : 1004 ( Monkey Banana Problem ) 
// File name    : main.cpp
// Author       : iCoding
// E-mail       : honi.linux@gmail.com
// Date & Time  : Thu Aug  9 12:13:15 2012


#include <iostream>
#include <stdio.h>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

/*************************************************************************************/
/* data */
#ifndef MAXN
#define MAXN 110
#endif
int iMap[MAXN*2][MAXN];

const int BLOCK = -1;

int n;

/*************************************************************************************/
/* procedure */
bool iCanGo(int x, int y)
{
    if (iMap[x][y] == -1)
    {
        return false;
    }
    else
    {
        return true;
    }
}

void iInit()
{
    for (int i = 0; i < MAXN * 2; i++)
    {
        for (int j = 0; j < MAXN; j++)
        {
            iMap[i][j] = BLOCK;
        }
    }

    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= i; j++)
        {
            scanf("%d", &iMap[i][j]);
        }
    }
    for (int i = n + 1; i < 2 * n; i++)
    {
        for (int j = 1; j <= (2 * n - i); j++)
        {
            scanf("%d", &iMap[i][j]);
        }
    }
}

void iDynamicProgramming()
{
    for (int i = (n * 2 - 2); i >= n; i--)
    {
        for (int j = 1; j <= (2 * n - i); j++)
        {
            int iMax = -1;
            if (iCanGo(i+1,j-1) && iMap[i+1][j-1] > iMax)
            {
                iMax = iMap[i+1][j-1];
            }

            if (iCanGo(i+1,j) && iMap[i+1][j] > iMax)
            {
                iMax = iMap[i+1][j];
            }

            iMap[i][j] = iMax + iMap[i][j];
        }
    }

    for (int i = n - 1; i >= 1; i--)
    {
        for (int j = 1; j <= i; j++)
        {
            if (iMap[i+1][j] > iMap[i+1][j+1])
            {
                iMap[i][j] += iMap[i+1][j];
            }
            else
            {
                iMap[i][j] += iMap[i+1][j+1];
            }
        }
    }
}

void iShowMap()
{
    for (int i = 0; i <= n * 2; i++)
    {
        for (int j = 0; j <= n; j++)
        {
            printf("%d ", iMap[i][j]);
        }
        printf("\n");
    }
}

void iShowResult()
{
    printf("%d\n", iMap[1][1]);
}

/*************************************************************************************/
/* main */
int main()
{
    int iT;
    scanf("%d", &iT);
    for (int iCaseCount = 1; iCaseCount <= iT; iCaseCount++)
    {
        printf("Case %d: ", iCaseCount);
        iInit();
        iDynamicProgramming();
        //iShowMap();
        iShowResult();

    }
    return 0;
}

// end 
// Code by Sublime text 2
// iCoding@CodeLab