Hopscotch（dfs）

Hopscotch
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4453 Accepted: 2973

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.

Source

```#include

#include

#include

#include

#include

using namespace std;int a[100010],mp[6][6],sum,ans;int dir[4][2]={
{-1,0},{1,0},{0,-1},{0,1}};void dfs(int x,int y,int m,int n)   //xy代表坐标，m代表目前所形成的数，n代表长度{
int i;
if(n==6)
{
a[ans++]=m;   //找到一种情况。
return;
}
m=m*10+mp[x][y];
for(i=0;i<4;i++)
{
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(xx>=0&&xx<5&&yy>=0&&yy<5)
{
dfs(xx,yy,m,n+1);
}
}}int main(){
int i,j;
ans=0;
for(i=0;i<5;i++)
for(j=0;j<5;j++)
{
scanf("%d",&mp[i][j]);
}
for(i=0;i<5;i++)
for(j=0;j<5;j++)
{
dfs(i,j,0,0);
}
sort(a,a+ans);
sum=1;
for(i=0;i

{
if(a[i]!=a[i+1])sum++;
}
printf("%d\n",sum);
return 0;}

```

#### 最新留言

[***.172.111.71]2022年05月22日 09时45分19秒

## 关于作者

喝酒易醉，品茶养心，人生如梦，品茶悟道，何以解忧？唯有杜康！
-- 愿君每日到此一游！