POJ 3050 Hopscotch

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2090		Accepted: 1500

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

题意：

在一个5X5 的格子里往四个方向跳，分别为上下左右，求共有几种不同字符串。

将字符串转换成整数，dfs + set 容器

代码如下：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<set>
#define For(i, a, b) for(int i = a; i < b; i++)
using namespace std;
set <int> S;
int Imap[6][6];
int dir[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
int cleck(int x, int y)
{
    return (x >= 0 && x < 5 && y >= 0 && y <5);
}

void dfs(int x, int y, int sum, int pos)
{
    if(pos == 6) {
        S.insert(sum);
        return;
    }
    sum = 10 * sum + Imap[x][y];
    For(i, 0, 4){
        int tx = x + dir[i][0];
        int ty = y + dir[i][1];
        if(cleck(tx, ty))
            dfs(tx, ty, sum, pos+1);
    }
}
int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d", &Imap[0][0])){
        For(i, 0, 5) For(j, 0, 5){
            if(i == 0 && j == 0) continue;
            scanf("%d", &Imap[i][j]);
        }
        For(i, 0, 5) For(j, 0, 5) dfs(i, j, 0, 0);
        printf("%d\n", S.size());
    }
    return 0;
}