1004 - Monkey Banana Problem

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You are in the world of mathematics to solve the great "MonkeyBanana Problem". It states that, a monkey enters into a diamond shaped twodimensional array and can jump in any of the adjacent cellsdown fromits current position (see figure). While
moving from one cell to another, themonkey eats all the bananas kept in that cell. The monkey enters into the arrayfrom the upper part and goes out through the lower part. Find the maximumnumber 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 be2*N - 1 rows. The i^th(1 ≤ i ≤ N) line of nextN lines contains exactlyinumbers. Then
there will beN - 1 lines. The j^th (1≤ j < N) line containsN - j integers. Each number isgreater than zero and less than 2¹⁵.

Output

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

Note

Dataset is huge, use faster I/O methods.

题目链接：http://lightoj.com/volume_showproblem.php?problem=1004

题目大意：求从上往下路径的最大权值。

解题思路：数字三角形的变形，把菱形分为上下两部分求即可。dp[i][j]表示路径到第i行第j个点的最大权值和。

首先输入正三角，j=1时：dp[i][j] = dp[i-1][j];

j=n时：dp[i][j] = dp[i-1][j-1];

j=2~n-1: dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);

然后输入倒三角，dp[i][j]=max(dp[i-1][j],dp[i-1][j+1]);

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int a[102][102],dp[102][102];
int main()
{
    int i,j,t,n,cnt=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            for(j=1;j<=i;j++)
            scanf("%d",&a[i][j]);
        dp[1][1]=a[1][1];
        for(i=2;i<=n;i++)
            for(j=1;j<=i;j++)
            {
                if(j==1)
                    dp[i][j]=a[i][j]+dp[i-1][j];
                else if(j==n)
                    dp[i][j]=a[i][j]+dp[i-1][j-1];
                else
                    dp[i][j]=a[i][j]+max(dp[i-1][j-1],dp[i-1][j]);
            }
        for(i=n-1;i>=1;i--)
            for(j=1;j<=i;j++)
                scanf("%d",&a[i][j]);
        for(i=n-1;i>=1;i--)
            for(j=1;j<=i;j++)
                dp[i][j]=a[i][j]+max(dp[i+1][j],dp[i+1][j+1]);
        printf("Case %d: %d\n",++cnt,dp[1][1]);
    }
    return 0;
}