POJ 3050：Hopscotch

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2506		Accepted: 1784

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.

这个题题意是给一个5*5的矩阵，你可以从任意一个起点走5步，每一步可以上下左右那么走，记录你走过的路径，即你脚下的位置的value，输出不同的路径个数。

自己想输出不同的值，用的vector。。。结果一看其他人用的set比我的方便好多，而且直接insert啊，set自动去重啊。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <queue>
#pragma warning(disable:4996)
using namespace std;

int value[7][7];
vector<int>road;

void dfs(int i,int j,int step,int test)
{
if(step==6)
{
test = test*10+value[i][j];
road.push_back(test);
return;
}
test = test*10+value[i][j];

if(i>1)
{
dfs(i-1,j,step+1,test);
}
if(j>1)
{
dfs(i,j-1,step+1,test);
}
if(i<5)
{
dfs(i+1,j,step+1,test);
}
if(j<5)
{
dfs(i,j+1,step+1,test);
}
}

int main()
{
int i,j;
for(i=1;i<=5;i++)
{
for(j=1;j<=5;j++)
{
cin>>value[i][j];
}
}
for(i=1;i<=5;i++)
{
for(j=1;j<=5;j++)
{
dfs(i,j,1,0);
}
}
sort(road.begin(), road.end());
vector<int>::iterator iter =unique(road.begin(),road.end());
road.erase(iter,road.end());

cout<<road.size()<<endl;
return 0;
}