Light oj-1004 - Monkey Banana Problem,数字三角形的变形版~

                                                                                                 1004 - Monkey Banana Problem

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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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 i numbers.
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)

说白了这道题就是两个数字三角形，一个正的，一个倒过来的；倒着的做法与数字三角形一样，只需要处理正的；

我们来看样例，把这个分为上下两部分，下面就是个倒的数字三角形，正的三角形其实与倒的处理方式类似，我们做数字三角形是从下往上推，或者从上往下推，如果两种都会的话这道题没什么问题了；

博主其实做这个以前都是从下往上推的，但这题分为两部分就不好处理了，所以我选择正三角形从上往下推，而倒三角形相当于一个正的三角形从下往上推，做法不就和纯的数字三角形一模一样了；若不懂来看代码；

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
//const int INF=0x3f3f3f3f;
const int N=200+10;//开二维数组足够了，没有用到空间压缩；
long long a[N][N];
int main()
{
    int t,n,i,j;
    memset(a,0,sizeof(a));
    scanf("%d",&t);
    int t1=t;
    while(t--)
    {
        scanf("%d",&n);
        for(i=1; i<=2*n-1; i++)
        {
            if(i<=n)
            {
                for(j=1; j<=i; j++)
                    scanf("%lld",&a[i][j]);//正三角形；
            }
            else
            {
                for(j=1; j<=2*n-i; j++)//倒三角；
                    scanf("%lld",&a[i][j]);
            }
        }
        printf("Case %d: ",t1-t);
        for(i=2; i<=n; i++)
            for(j=1; j<=i; j++)
                a[i][j]+=max(a[i-1][j-1],a[i-1][j]);//从上往下，状态转移方程；
        for(i=n+1; i<=2*n-1; i++)
            for(j=1; j<=2*n-i; j++)
                a[i][j]+=max(a[i-1][j],a[i-1][j+1]);//状态转移方程几乎一样
        printf("%lld\n",a[2*n-1][1]);注意格式，博主用I64dPEI
    }
    return 0;
}