poj:3050 Hopscotch (Dfs+set)

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4071		Accepted: 2708

Description The cows play the child's game of hopscotch in a non-traditional. 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 has leading zeroes like 000201).

Determine the count of the number of distinct integers that can is created in this manner.

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

Output * Line 1:the number of distinct integers so 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 is possible.

Source

Usaco 2005 November Bronze

To give you a 5*5 number table, ask a number of 6-step permutations generated from any digit.

Problem-Solving ideas: Dfs for each element, putting the results of each DFS into set to prevent duplication. Save 6 Steps: 1 2 1 2 1 1 This path becomes the number 121211.

The code is as follows:

#include <cstdio>
#include <queue>
#include <set>
#include <iostream>
# Include <string>
using namespace std;
int map[5][5];
int dirx[4]={-1,1,0,0};
int diry[4]={0,0,-1,1};
set<int> s;
int judge (int x,int y)
{
	if (x>=0&&x<5&&y>=0&&y<5)
	{
		return 1;
	}
	else
	{
		return 0;
	}
}
void Dfs (int hang,int lie,int sum,int depth)
{
	if (depth==6)
	{
		s.insert (sum);//If the condition already exists in it, Then the size of s does not change to 
		return;
	}
	for (int i=0;i<4;i++)
	{
		int nextx=dirx[i]+hang;
		int Nexty=diry[i]+lie;
		if (judge (nextx,nexty))
		{
			dfs (nextx,nexty,sum*10+map[hang][lie],depth+1);//The third parameter converts the path to a number 
		}
	}
}
int main ()
{
	s.clear ();
	for (int i=0;i<5;i++)
	{
		for (int j=0;j<5;j++)
		{
			scanf ("%d", &map[i][j]);
		}
	}
	for (int i=0;i<5;i++)
	{for
		(int j=0;j<5;j++)
		{
			dfs (i,j,0,0);
		}
	}
	printf ("%d\n", S.size ());
	return 0;
}