poj3050(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.

以所有的点为起点进行深度为6的DFS，将得到的数放到STL的set中，最后统计共有多少个数。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};
const int maxn = 10000010;

int s[5][5];
set<int> m;

bool range(int x, int y)
{
    return x >= 0 && x < 5 && y >= 0 && y < 5;
}

void dfs(int x, int y, int res, int cur)
{
    if (cur == 5)
    {
        m.insert(res);
        return;
    }
    for (int i = 0; i < 4; ++i)
    {
        int tx = x + dx[i], ty = y + dy[i];
        if (range(tx, ty))
        {
            int tmp = res * 10 + s[tx][ty];
            dfs(tx, ty, tmp, cur+1);
        }
    }
}

int main()
{
    for (int i = 0; i < 5; ++i)
        for (int j = 0; j < 5; ++j)
            scanf("%d", &s[i][j]);
    for (int i = 0; i < 5; ++i)
        for (int j = 0; j < 5; ++j)
            dfs(i, j, s[i][j], 0);
    int ans = 0;
    for (set<int>::iterator it = m.begin(); it != m.end(); ++it)
        ans++;
    cout << ans << endl;
    return 0;
}