HDU1208：Pascal's Travels（dp）

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解决一个游戏板上的路径寻找问题，玩家需要从左上角移动到右下角，每个方格内的数字决定了下一步的大小。使用动态规划算法来计算所有可能的路径数量。

Pascal's Travels

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2318 Accepted Submission(s): 1090

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

题意：每格数字代表能向下或向右移动的步数，求左上到右下有多少条路径。

思路：略。

# include <stdio.h>
# include <string.h>
int main()
{
    int n;
    char a[35][35];
    long long dp[35][35];
    while(~scanf("%d",&n),n!=-1)
    {
        memset(dp, 0, sizeof(dp));
        for(int i=0; i<n; ++i)
            scanf("%s",&a[i]);
        if(a[0][0] == '0')
        {
            puts("0");
            continue;
        }
        else
            dp[0][0] = 1;
        for(int i=0; i<n; ++i)
            for(int j=0; j<n; ++j)
            {
                if(dp[i][j] && a[i][j]-'0')
                {
                    if(j+a[i][j]-'0' < n)
                        dp[i][j+a[i][j]-'0'] += dp[i][j];
                    if(i+a[i][j]-'0' < n)
                        dp[i+a[i][j]-'0'][j] += dp[i][j];
                }
            }
        printf("%lld\n",dp[n-1][n-1]);
    }
    return 0;
}