Pascal's Travels

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

题意：给定一个二维图，给出图中每个点到其他点的最大步数，求从最左上出发到最右下的的步数

题解：定义一个二维数组p[i][j],p[0][0]=1;设当前点能走的最大步数为k

p[i+k][j]+=p[i][j];p[i][j+k]+=p[i][j];

AC代码：

#include<stdio.h>
#include<string.h>
__int64 p[50][50];
char ms[50][50];
int main()
{
    __int64 i,n,j,k;
 while(scanf("%I64d",&n)&&n!=-1)
 {
  memset(p,0,sizeof(p));
  p[0][0]=1;
  for(i=0;i<n;i++)
  {
   scanf("%s",ms[i]);
  }
  for(i=0;i<n;i++)
  {
   for(j=0;j<n;j++)
   {
    k=ms[i][j]-'0';
                 if(k==0) continue;
     if(j+k<n) p[i][j+k]+=p[i][j];
     if(i+k<n) p[i+k][j]+=p[i][j];
   }
  }
  printf("%I64d\n",p[n-1][n-1]);
 }
 return 0;
}