HDU 1208 Pascal's Travels 经典 跳格子的方案数 （dp或者记忆化搜索）

Pascal's Travels

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

分析：

思路1：

dp[i][j]：表示从1，1到i，j的方案数目

注意初始化dp[1][1]=1

状态转移方程：

 if(i+a[i][j]<=n)
     dp[i+a[i][j]][j]+=dp[i][j];
 if(j+a[i][j]<=n)
     dp[i][j+a[i][j]]+=dp[i][j];

下一个状态有当前状态决定

code：

#include <iostream>
#include <cstdio>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<math.h>
#include<memory>
#include<queue>
#include<vector>
using namespace std;
#define max_v 40
__int64 dp[max_v][max_v];//dp[i][j] 从1，1到i，j的方案数
int a[max_v][max_v];
char s[max_v];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==-)
            break;
        for(int i=;i<=n;i++)
        {
            scanf("%s",s);
            int l=strlen(s);
            int k=;
            for(int j=;j<l;j++)
            {
                a[i][k++]=s[j]-'';
            }
        }
        memset(dp,,sizeof(dp));
        dp[][]=;
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=n;j++)
            {
                if(a[i][j]==)
                    continue;
                if(i+a[i][j]<=n)
                    dp[i+a[i][j]][j]+=dp[i][j];
                if(j+a[i][j]<=n)
                    dp[i][j+a[i][j]]+=dp[i][j];
            }
        }
        printf("%I64d\n",dp[n][n]);
    }
    return ;
}

思路二：

记忆化搜索

注意这里dp的含义和上面dp的含义不同

这里的dp：dp[i][j]代表从i，j出发的方案数

#include <iostream>
#include <cstdio>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<math.h>
#include<memory>
#include<queue>
#include<vector>
using namespace std;
#define max_v 40
__int64 dp[max_v][max_v];//dp[i][j] 从i，j出发的方案数
int a[max_v][max_v];
char s[max_v];
int n;
__int64 dfs(int x,int y)
{
    if(dp[x][y]>)//记忆化搜索
        return dp[x][y];
    if(x==n&&y==n)//搜到终点
        return ;
    int xx,yy;
    if(a[x][y]==)//不能跳的点
        return ;
    for(int k=;k<=;k++)//两个方向，下或右
    {
        if(k==)
        {
            xx=x+a[x][y];
            yy=y;
        }else
        {
            xx=x;
            yy=y+a[x][y];
        }
        if(xx<=n&&yy<=n)//避免越界
            dp[x][y]+=dfs(xx,yy);
    }
    return dp[x][y];
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(n==-)
            break;
        for(int i=;i<=n;i++)
        {
            scanf("%s",s);
            int l=strlen(s);
            int k=;
            for(int j=;j<l;j++)
            {
                a[i][k++]=s[j]-'';
            }
        }
        memset(dp,,sizeof(dp));
        printf("%I64d\n",dfs(,));//从1，1开始搜
    }
    return ;
}