poj 3050 Hopscotch(暴力dfs)

问题描述

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.

输入

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

输出

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

样例输入

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

样例输出

15

开始看成了走5步结果WA了一发。。。
暴力dfs，直接一个一个的枚举，然后放入集合se里面，自动去重。最后再统计下se里面元素的个数即可。

代码如下：

#include <iostream>
#include <cstdio>
#include <set>
using namespace std;

int ma[6][6];
int num[10];
set<int> se;

int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};

void dfs(int x,int y,int deep)
{
    if(x>5||y>5||x<=0||y<=0)
        return ;
    if(deep==6)
    {
        int k=0;
        for(int i=0;i<deep;i++)
        {
            k=k*10+num[i];
        }
        se.insert(k);
        return ;
    }
    num[deep]=ma[x][y];
    for(int i=0;i<4;i++)
    {
        dfs(x+dx[i],y+dy[i],deep+1);
    }
}

int main()
{
    for(int i=1;i<=5;i++)
        for(int j=1;j<=5;j++)
        scanf("%d",&ma[i][j]);

    for(int i=1;i<=5;i++)
        for(int j=1;j<=5;j++)
        dfs(i,j,0);
        printf("%d\n",se.size());
}