hdu 1208 Pascal's Travels (DP记忆化搜索)

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本文介绍了一个用于计算在特定游戏板上从左上角到右下角路径数量的方法，通过递归和动态规划解决。适用于4x4至34x34的游戏板。

Pascal's Travels

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1115 Accepted Submission(s): 488

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

Sample Output

  
  
   
   3
0
7


   
   
    
    
     
     Hint
    
    Hint
   
   
   
    
Brute force methods examining every path will likely exceed the allotted time limit. 
64-bit integer values are available as "__int64" values using the Visual C/C++ or "long long" values 
using GNU C/C++ or "int64" values using Free Pascal compilers.

Source

Mid-Central USA 2005

Recommend

mcqsmall

思路:没啥好说的，目标点记为1，从开始点开始，走右和下，和加起来即可，具体代码一看便知。

#include<iostream>
#include<cstring>
using namespace std;
const int mm=37;
long long dp[mm][mm];
char f[mm][mm];
bool vis[mm][mm];
int m;
long long DP(int x,int y)
{
  if(x<0||x>=m||y<0||y>=m)return 0;
  if(vis[x][y])return dp[x][y];
  if(f[x][y]=='0')return 0;
  dp[x][y]=DP(x+f[x][y]-'0',y)+DP(x,y+f[x][y]-'0');
  vis[x][y]=1;
  return dp[x][y];
}
int main()
{
  while(cin>>m)
  { if(m==-1)break;
   memset(dp,0,sizeof(dp));
   memset(vis,0,sizeof(vis));
    for(int i=0;i<m;i++)
      {
         cin>>f[i];
      }
      dp[m-1][m-1]=1;vis[m-1][m-1]=1;
    cout<<DP(0,0)<<"\n";
  }
}