HDU1208 Pascal's Travels(DP)An n x n game board is populated

题目：

Pascal's Travels

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2296 Accepted Submission(s): 1079
**
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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.
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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

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思路：

我先说一下这个题的意思，题目的意思是给了一张图，要从左上角走到右下角，每一个小格的数字代表，这个人在这里走下一步要走几格，而且只能向下或者向右走，以第一个样例来说,第一个数字是2，那么他能向右或者向下走，向右走到了3这个数字，于是这一条路就是（2->3->1->0）,第二条路是向下走，则路径是(2->1->3->0),还有一条路，因为走到数字1时有两种情况(2->1->2->1->0),所以第一个样例的可能性有三种。

我们用dp[i][j]来表示要到达第i行和到达第j列所要走的步数。

代码：

#include <stdio.h>
#include <string.h>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
long long dp[40][40];
char map[40][40];
int main()
{
    int n;
    while(scanf("%d",&n)&&n!=-1)
    {
        for(int i=0; i<n; i++)
        {
            scanf("%s",map[i]);
            for(int j=0; j<n; j++)
                map[i][j]-='0';//以字符形式输入，在这里变成整数值
        }
        mem(dp,0);
        dp[0][0]=1;//在第0行第0列他只能走一步
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            {
                if(map[i][j]==0||dp[i][j]==0)
                    continue;
                if(i+map[i][j]<n)//横着走，检查是否越界
                    dp[i+map[i][j]][j]+=dp[i][j];//加上他当前的步数的可能性
                if(j+map[i][j]<n)//竖着走，检查是否越界
                    dp[i][j+map[i][j]]+=dp[i][j];
            }
        printf("%lld\n",dp[n-1][n-1]);//数组的行和列是从0开始的，所以这里要减1
    }
    return 0;
}