POJ 3050 Hopscotch (穷竭搜索)

题目链接：http://poj.org/problem?id=3050

题面：

Hopscotch

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.

Source

USACO 2005 November Bronze

题目大意：

一只牛在5*5的方格内可以从任意点开始跳，每次可以往4个方向跳，共跳5次，根据走过的方格，形成一个长度为6的序列，问这样的序列有多少种？（方格可以重复跳。）

解题：

题目挂的时候写的是穷竭搜索，看到图这么小，估计就是暴搜的意思吧。枚举从每个点开始跳，跳到5步后停止，并记录此时形成的路径，用set排重计数即可。

代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <set>
#include <vector>
using namespace std;
char map[6][6],path[10];
//四个方向
int dir[4][2]={0,1,-1,0,0,-1,1,0};
//计数
set <string> cnt;
string tmp;
//判断是否方格内
bool inside(int x,int y)
{
	if(x>=1&&x<=5&&y>=1&&y<=5)
		return true;
	else
		return false;
}
//深搜
void dfs(int x,int y,int step)
{
	int tx,ty;
	path[step]=map[x][y];
	if(step==5)
	{
		//记录路径
	   tmp="";
       for(int i=0;i<=5;i++)
		   tmp+=path[i];
	   cnt.insert(tmp);
	   return;
	}
	//往四个方向跳
	for(int i=0;i<4;i++)
	{
		tx=x+dir[i][0];
		ty=y+dir[i][1];
		if(inside(tx,ty))
			dfs(tx,ty,step+1);
	}
}
int main()
{
	//读入
    for(int i=1;i<=5;i++)
		for(int j=1;j<=5;j++)
			scanf(" %c",&map[i][j]);
	//枚举从每个点开始
	for(int i=1;i<=5;i++)
		for(int j=1;j<=5;j++)
			dfs(i,j,0);
	//输出
	printf("%d\n",cnt.size());
	return 0;
}

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