POJ 3050——Hopscotch(dfs,爆搜)

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本文介绍了一种基于矩阵的游戏——Hopscotch。通过从任意起点出发，向四个方向移动五次来形成六位数，并统计所有可能的不同数值。文章提供了完整的实现代码，采用递归深度优先搜索的方法进行遍历。

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题目：

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3239		Accepted: 2245

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

Sample Output

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：给个矩阵，从任意起点出发。向左向右向上向下看能组成多少个六位数，可以有前导零。

思路：直接暴力搜索即可

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <queue>
#include <cctype>
#include <string>
#include <map>
#include <set>
#include <iostream>
using namespace std;

#define INF 1e9+7
#define MAXN 100010

int a[5][5];

int shift(int temp[])
{
    int s=0;
    for(int k=0;k<6;k++)
    {
        s=s*10+temp[k];
    }
    return s;
}

set <int> res;
int temp[6];
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};

void dfs(int x,int y,int cur)
{
    if(cur==6)
    {
        int ans=shift(temp);
        res.insert(ans);
        return;
    }
    temp[cur]=a[x][y];
    for(int i=0;i<4;i++)
    {
        int ex=x+dx[i];
        int ey=y+dy[i];
        if(ex>=0&&ex<5&&ey>=0&&ey<5)
        {
            dfs(ex,ey,cur+1);
        }
    }
}

int main()
{
    for(int i=0;i<5;i++)
        for(int j=0;j<5;j++)
            scanf("%d",&a[i][j]);
    for(int i=0;i<5;i++)
        for(int j=0;j<5;j++)
        {
            dfs(i,j,0);
        }
    printf("%d\n",(int)res.size());
}