zoukankan      html  css  js  c++  java
  • HDU 1208 Pascal's Travels

    Pascal's Travels

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 1607    Accepted Submission(s): 696


    Problem Description
    An n x n game board is populated with integers, one nonnegative integer per square. The goal is to travel along any legitimate path from the upper left corner to the lower right corner of the board. The integer in any one square dictates how large a step away from that location must be. If the step size would advance travel off the game board, then a step in that particular direction is forbidden. All steps must be either to the right or toward the bottom. Note that a 0 is a dead end which prevents any further progress.


    Consider the 4 x 4 board shown in Figure 1, where the solid circle identifies the start position and the dashed circle identifies the target. Figure 2 shows the three paths from the start to the target, with the irrelevant numbers in each removed.


    Figure 1


    Figure 2
     
    Input
    The input contains data for one to thirty boards, followed by a final line containing only the integer -1. The data for a board starts with a line containing a single positive integer n, 4 <= n <= 34, which is the number of rows in this board. This is followed by n rows of data. Each row contains n single digits, 0-9, with no spaces between them.
     
    Output
    The output consists of one line for each board, containing a single integer, which is the number of paths from the upper left corner to the lower right corner. There will be fewer than 2^63 paths for any board.
     
    Sample Input
    4
    2331
    1213
    1231
    3110
    4
    3332
    1213
    1232
    2120
    5
    11101
    01111
    11111
    11101
    11101
    -1
     
    Sample Output
    3
    0
    7
    讲解:和这道题比较类似 How many ways
    
    
     1 #include<iostream>
     2 #include<algorithm>
     3 #include<queue>
     4 #include<cstring>
     5 using namespace std;
     6 int  mapp[100][100];
     7 int n;
     8 long long dp[40][40];
     9 long long dfs(int x,int y)
    10 {
    11         int xx,yy;
    12         if(x==n && y==n)
    13                return 1;
    14         if(mapp[y][x]==0 || dp[y][x]>0)
    15              return dp[y][x];
    16         for(int i=0;i<2;i++)
    17           {
    18               if(i==0)
    19               {
    20                   xx=x+mapp[y][x]; yy=y;
    21               }
    22               else
    23               {
    24                   xx=x;yy=y+mapp[y][x];
    25               }
    26               if(xx<=n && yy<=n)
    27                 dp[y][x]+=dfs(xx,yy);
    28           }
    29      return dp[y][x];
    30 }
    31 int main()
    32 {
    33     char s;
    34     while(cin>>n && n>-1)
    35     {
    36         memset(mapp,0,sizeof(mapp));
    37         memset(dp,0,sizeof(dp));
    38         for(int i=1; i<=n; i++)
    39             for(int j=1; j<=n; j++)
    40             {
    41                cin>>s;
    42                mapp[i][j]=s-'0';
    43             }
    44         long long sum=dfs(1,1);
    45         cout<<sum<<endl;
    46     }
    47     return 0;
    48 }
    
    
    
    
    
     
  • 相关阅读:
    [51nod 1129] 字符串最大值(kmp)
    P3391 【模板】文艺平衡树(Splay)
    次大公约数
    青蛙的约会
    [HNOI2002]营业额统计
    GYM 100741A Queries
    P3370 【模板】字符串哈希
    P3369 【模板】普通平衡树(Treap/SBT)
    05:LGTB 与偶数
    简单计算器
  • 原文地址:https://www.cnblogs.com/lovychen/p/3725541.html
Copyright © 2011-2022 走看看