poj比赛题1006

 

Cow Solitaire

Time Limit:3000MS  Memory Limit:65536K
Total Submit:14 Accepted:8

Description

Late summer on the farm is a slow time, very slow. Betsy has little to do but play cow solitaire. For self-evident reasons, cow solitaire is not so challenging as any number of solitaire games played by humans.

Cow solitaire is played using an N x N (3 <= N <= 7) grid of ordinary playing cards with four suits (Clubs, Diamonds, Hearts, and Spaces) of 13 cards (Ace, 2, 3, 4, ..., 10, Jack, Queen, King). Cards are named with two characters: their value (A, 2, 3, 4, ..., 9, T, J, Q, K) followed by their suit (C, D, H, S). Below is a typical grid when N = 4:

8S AD 3C AC (Eight of Spades, Ace of Diamonds, etc.)
8C 4H QD QS
5D 9H KC 7H
TC QC AS 2D

To play this solitaire game, Betsy starts in the lower left corner (TC) and proceeds using exactly 2*N-2 moves of 'right' or 'up' to the upper right corner. Along the way, she accumulates points for each card (Ace is worth 1 point, 2 is worth 2 points, ..., 9 is worth 9 points, T is worth 10 points, J is 11, Q is 12, and K is 13) she traverses. Her goal is to amass the highest score.

If Betsy's path was TC-QC-AS-2D-7H-QS-AC, her score would be 10+12+1+2+7+12+1=45. Had she taken the left side then top (TC-5D-8C-8S-AD-3C-AC), her score would be 10+5+8+8+1+3+1=36, not as good as the other route. The best score for this grid is 69 points (TC-QC-9H-KC-QD-QS-AC => 10+12+9+13+12+12+1).Betsy wants to know the best score she can achieve. One of the geek cows once told her something about "working from the end back to the beginning," but she didn't understand what they meant.

Input

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 lists the cards on row i (row 1 is the top row) using N space-separated card names arranged in the obvious order.

Output

* Line 1: A single line with an integer that is the best possible score Betsy can achieve.

Sample Input

48S AD 3C AC8C 4H QD QS5D 9H KC 7HTC QC AS 2D

Sample Output

69

昨天迎来了我们队的第一次内部比赛,很郁闷,自己居然连一道题也没能做出,哎,不

愦为一个超级大菜鸟,老实说,对于平时松散习惯了的我,对于成为ACM的正式队员并没

有很强的欲望,但是,我不想一直做那菜鸟中的菜鸟,不甘心啦,,,,,,

在比赛中我犯了一个错误,那就是在一棵树上吊死了,整整4个小时,我知识把精力放

在了一道题上面,而那道题却是其中最难的一道,寒,,,,,,,

对于本题,与我当天上午做的一道题很类似,但是,还是让我在晚上折腾了很久才做出,

用了很多种错误的方法,最终还是回归到了动态规划的思想.

正确代码如下:

#include<iostream>
using namespace std;
int main()
{
  char c,d;
  int i,j,m,a[8][8];
  cin>>m;
  for(i=0;i<m;i++)
     {
      for(j=0;j<m;j++)
       {
          cin>>c>>d;
          if(c=='A')   a[i][j]=1;
          else if(c=='T')  a[i][j]=10;
          else if(c=='J')  a[i][j]=11;
          else if(c=='Q')  a[i][j]=12;
          else if(c=='K')  a[i][j]=13;
          else
              a[i][j]=c-48;
       }
       }
    int num[8]={0};
    for(i=0;i<m;i++)
    {
      num[m-1]+=a[i][m-1];    
      for(j=m-2;j>=0;j--)
       {
       num[j]+=a[i][j];     
      if(num[j]<num[j+1]+a[i][j])   
          num[j]=num[j+1]+a[i][j];    
          }   
       }
    for(i=1;i<m;i++)
       for(j=0;j<i;j++)
         num[i]+=a[m-1][j];
 int max=0;
 for(i=0;i<m;i++)
    if(max<num[i])
        max=num[i];
    cout<<max;                        
 return 0;
}