POJ-3050 Hopscotch

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本文介绍了一种基于深度优先搜索(DFS)的算法解决方案，用于解决POJ 3050 Hopscotch问题。该问题要求在一个5x5数字网格中通过五次移动形成不同的六位数，并统计这些数目的总数。文章提供的代码使用了递归深搜方法，并借助set集合去除重复计数。

题目链接：

http://poj.org/problem?id=3050

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6053		Accepted: 3966

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

Sample Output

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.

题目分析：

利用深搜，但是要注意形成的六位数要考虑相同，所以可以利用set集合存储，最后输出size就可以

#include<iostream>
#include<set>
using namespace std;
int a[5][5];
set<int>s;
int moveX[]={-1,1,0,0};
int moveY[]={0,0,-1,1};
void dfs(int x,int y,int k,int num)
{
	if(k==6){
		s.insert(num);
		return;
	}
	else{
		
		for(int i=0;i<4;i++)
		{
			int nx = x + moveX[i];
			int ny = y + moveY[i];
		    if(nx>=0&&nx<=4&&ny>=0&&ny<=4){
				k++;
				dfs(nx,ny,k,num*10+a[nx][ny]);
				k--;
			}
		}			
	}
}
int main(void)
{
	for(int i = 0;i<5;i++)
	{
		for(int j=0;j<5;j++)
		{
			cin>>a[i][j];
		}
	}
	for(int i=0;i<5;i++)
	{
		for(int j=0;j<5;j++)
		{
			dfs(i,j,1,a[i][j]);
		}
	}
	cout<<s.size()<<endl;
	return 0;
}