poj 3050 Hopscotch DFS+暴力搜索+set容器

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2774		Accepted: 1940

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.

1.对于上面的每一组数，刚开始是以字符串处理的，，后来看了一下别人的博客，

才发现直接十进制数的大小不同可以省掉很多麻烦

2.重点在于set集合的使用，，其有去重功能，，貌似白书上介绍过

#include <iostream>

#include<cstdio>

#include<set>

#include<cstring>

using namespace std;

int a[7][7];

int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};

set<int> q;

int dfs(int step,int x,int y,int v)

{

     if(step==6)

         q.insert(v);

     else

            for(int i=0;i<=3;i++)

            {

                 int kx=x+dx[i];

                 int ky=y+dy[i];

                 if(kx>=1&&kx<=5&&ky>=1&&ky<=5)

                   dfs(step+1,kx,ky,v*10+a[kx][ky]);

            }

    return 0;

}

int main()

{

    for(int i=1;i<=5;i++)

        for(int j=1;j<=5;j++)

          scanf("%d",&a[i][j]);

    for(int i=1;i<=5;i++)

        for(int j=1;j<=5;j++)

             dfs(1,i,j,a[i][j]);

    printf("%d\n",q.size());

    return 0;

}