F - Monkey Banana Problem

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Description

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

2

4

7

6 4

2 5 10

9 8 12 2

2 12 7

8 2

10

2

1

2 3

1

Sample Output

Case 1: 63

Case 2: 5

//题意第一行是 T 组测试数据，第二行 n 然后是 2*n-1 行的数据

从最上面那里出发，只能选做下走或者右下走，走到最下面那个位置，求路径上最大的和。

//很简单的 dp 注意一些特殊的情况就行

//72ms

 1 #include <stdio.h>
 2 #include <string.h>
 3
 4 int num[210][105];
 5 int dp[210][105];
 6
 7 int max(int a ,int b)
 8 {return  a>b?a:b;}
 9
10 int main()
11 {
12     int T,Case=0;
13     int i,j;
14     scanf("%d",&T);
15     while (T--)
16     {
17         int n;
18         scanf("%d",&n);
19         for (i=1;i<=2*n-1;i++)
20         {
21             if (i<=n)
22                 for (j=1;j<=i;j++)
23                     scanf("%d",&num[i][j]);
24             else
25                 for (j=1;j<=2*n-i;j++)
26                     scanf("%d",&num[i][j]);
27         }
28
29         memset(dp,0,sizeof(dp));
30         for (i=2*n-1;i>=1;i--)
31         {
32             if (i<n)
33             {
34                 for (j=1;j<=i;j++)
35                     dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+num[i][j];
36             }
37             else
38             {
39                 for (j=1;j<=2*n-i;j++)
40                 {
41                     if (j==1)
42                         dp[i][j]=dp[i+1][j]+num[i+1][j];
43                     if (j!=1&&j==2*n-i)
44                         dp[i][j]=dp[i+1][j-1]+num[i][j];
45
46                     dp[i][j]=max(dp[i+1][j-1],dp[i+1][j])+num[i][j];
47                 }
48             }
49         }
50         printf("Case %d: ",++Case);
51         printf("%d\n",dp[1][1]);
52     }
53     return 0;
54 }