POJ 3050 Hopscotch

Description

The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows create a 5x5 rectilinear grid of digits parallel to the x and y axes.

They then adroitly hop onto any digit in the grid and hop forward, backward, right, or left (never diagonally) to another digit in the grid. They hop again (same rules) to a digit (potentially a digit already visited).

With a total of five intra-grid hops, their hops create a six-digit integer (which might have leading zeroes like 000201).

Determine the count of the number of distinct integers that can be created in this manner.
Input

* Lines 1..5: The grid, five integers per line
Output

* Line 1: The number of distinct integers that can be constructed
Sample Input

1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 2 1
1 1 1 1 1

Sample Output

15

Hint

OUTPUT DETAILS:

111111, 111112, 111121, 111211, 111212, 112111, 112121, 121111, 121112, 121211, 121212, 211111, 211121, 212111, and 212121 can be constructed. No other values are possible.

题目大意：给一个5X5矩阵，问在任意一点出发，向四周走5步有多少种不同的走法；

题解：用set存储序列，每一个点进行dfs，得到的序列转化为一个数字，重复的序列就不会被计入，然后将容器大小输出即可；

AC代码：

#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std ;
int dp[6][6] ,vis[6][6],ans;
int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
set<int>ma;
void dfs(int x ,int y , int t ,int sum)
{
if(t==5)
{
ma.insert(sum);
return;
}
for(int i = 0 ; i < 4 ; i++)
{
int dx = x + dir[i][0];
int dy = y + dir[i][1];
if(dx>=0&&dx<5&&dy>=0&&dy<5)
{
dfs(dx,dy,t+1,dp[dx][dy]+sum*10);//*把当前的序列变为一个整数；
}
}
}

int main()
{
for(int i = 0 ; i<5;i++)
{
for(int j = 0 ;j<5;j++)
{
cin>>dp[i][j];
}
}
for(int i = 0 ; i<5;i++)
{
for(int j = 0 ;j<5;j++)
{
dfs(i,j,0,dp[i][j]);
}
}
cout<<ma.size()<<endl;
}