挑战2.1 Hopscotch（POJ 3050）

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3050

Appoint description: System Crawler  (2015-03-18)

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 11 1 1 1 11 1 1 1 11 1 1 2 11 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：

这道题是说，给你一个5*5的由正整数构成的矩阵，然后，能够在这个矩阵中构造出多少个不同的6位数字。

解题思路：

一看就是dfs题，这里用到了stl，因为是判断不能重复的数字的个数，所以用到set来处理下，最后直接输出s.size（）就OK了。

代码：

# include<cstdio># include<iostream># include<cstring># include<set>using namespace std;int grid[10][10];int next[4][2] = {{1,0},{0,-1},{-1,0},{0,1}};set<int>s;void dfs ( int x,int y,int step,int res ){    if ( step==6 )        {            s.insert(res);            return;        }    res = res*10+grid[x][y];    for ( int i = 0;i < 4;i++ )    {        int n_x = x+next[i][0];        int n_y = y+next[i][1];        if ( n_x >=0&&n_x < 5&&n_y>=0&&n_y<5 )        {            dfs(n_x,n_y,step+1,res);        }    }    return;}int main(void){    s.clear();    for ( int i = 0;i < 5;i++ )    {        for ( int j = 0;j < 5;j++ )        {            cin>>grid[i][j];        }    }    for ( int i = 0;i < 5;i++ )    {        for ( int j = 0;j < 5;j++ )        {            dfs( i,j,0,0 );        }    }    int ans = s.size();    cout<<ans<<endl;    return 0;}