hdu 1208（记忆化搜索）

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

记忆化搜索，刚开始直接暴力搜索tle了，所以还是乖乖dp吧，dp值就代表该点到终点的选择数。（记忆化搜索的代码其实是很好写的，稍微多做几题自然能掌握==），看代码吧：

#include <iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;
long long s;
int f[2][2]={{1,0},{0,1}},n;
long long dp[50][50];
char a[50][50];
long long  dfs(int i,int j)
{
    int flag=0;long long s=0;
    if(dp[i][j]!=-1)return dp[i][j];
    if(i==n-1&&j==n-1)return 1;
    if(a[i][j]=='0')
    {
            dp[i][j]=0;
            return 0;
    }
        for(int q=0;q<2;q++)
    {
        int X=(a[i][j]-'0')*f[q][0]+i,Y=(a[i][j]-'0')*f[q][1]+j;
        if(X>=0&&X<n&&Y>=0&&Y<n)
        {
            flag=1;
            s=s+dfs(X,Y);
        }
    }
    if(flag==0)
    {
        dp[i][j]=0;
        return dp[i][j];
    }else
    {
        dp[i][j]=s;return dp[i][j];
    }
}
int main()
{
    while(scanf("%d",&n)!=-1&&n!=-1)
    {
        s=0;
        for(int i=0;i<n;i++)
            scanf("%s",a[i]);
        memset(dp,-1,sizeof(dp));
        cout<<dfs(0,0)<<endl;
    }
    return 0;
}