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:
    [转] Linux 最大进程数, unable to create new native thread问题
    [转] Maven 从命令行获取项目的版本号
    [转]【JVM】调优笔记2-----JVM在JDK1.8以后的新特性以及VisualVM的安装使用
    DISCUZ 自定义模板
    Linux系统性能统计工具Sar和实时系统性能监控脚本
    shell脚本常规技巧
    Java中文编码小结
    json-smart 使用示例(推荐fastjson)
    HBase Java简单示例
    Ehcache BigMemory: 摆脱GC困扰
  • 原文地址:https://www.cnblogs.com/lovychen/p/3725541.html
Copyright © 2011-2022 走看看