poj 3050 Hopscotch（暴力+dfs）

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.

题意：从一点开始往上下左右进行跳跃5次，问最后能得到多少个不同的六位数。

思路：从数组的不同点进行dfs，当搜了6次且此时ans的bool值为0，把其变为1，tmp++，打印最后tmp的值。

AC代码：

 1 #include <iostream>
 2 #include<cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 #include<map>
 6 #include<algorithm>
 7 using namespace std;
 8 bool mp[1000000];
 9 int a[5][5],t[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
10 int ans;
11 int tmp;
12 void dfs(int x,int y,int dep)
13 {
14     int tx,ty;
15     ans=ans*10+a[x][y];
16     if(dep==6)
17     {
18         if(!mp[ans])
19         {
20             mp[ans]=1;
21             tmp++;
22         }
23         return ;
24     }
25     for(int i=0; i<4; i++)
26     {
27         tx=x+t[i][0],ty=y+t[i][1];
28         if(tx<0||ty<0||tx>=5||ty>=5)
29             continue;
30         else
31         {
32             dfs(tx,ty,dep+1);
33             ans=ans/10;
34         }
35     }
36 }
37 int main()
38 {
39     for(int i=0; i<5; i++)
40         for(int j=0; j<5; j++)
41             scanf("%d",&a[i][j]);
42     tmp=0;
43     for(int i=0; i<5; i++)
44         for(int j=0; j<5; j++)
45         {
46                 ans=0;
47                 dfs(i,j,1);
48         }
49     printf("%d\n",tmp);
50     return 0;
51 }

View Code

 

转载于:https://www.cnblogs.com/wang-ya-wei/p/6212863.html