POJ 3050 Hopscotch

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3372		Accepted: 2325

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

 

 

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
#include <string>
#include <cstring>
#include <queue>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
#define IOS ios_base::sync_with_stdio(false)
#define TIE std::cin.tie(0)
#define MIN2(a,b) (a<b?a:b)
#define MIN3(a,b) (a<b?(a<c?a:c):(b<c?b:c))
#define MAX2(a,b) (a>b?a:b)
#define MAX3(a,b,c)  (a>b?(a>c?a:c):(b>c?b:c))
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const double PI = 4.0*atan(1.0);

int ans, board[10][10], a[10];
int dx[4] = { 1, -1, 0, 0 }, dy[4] = { 0, 0, 1, -1 };
map<int, int> ma;
void dfs(int x, int y,int t)
{
    a[t] = board[x][y];
    if (t == 5){
        int tmp = 0;
        for (int i = 0; i < 6; i++){
            tmp *= 10; tmp += a[i];
        }
        if (!ma.count(tmp)){ ma[tmp] = 1; ans++; }
        return;
    }
    for (int i = 0; i < 4; i++){
        int nx = x + dx[i], ny = y + dy[i];
        if (0 <= nx && nx < 5 && 0 <= ny && ny < 5)
            dfs(nx, ny, t + 1);
    }
}
int main()
{
    for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++)
            scanf("%d", &board[i][j]);
    ans = 0;
    for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++)
            dfs(i, j, 0);
    printf("%d
", ans);
}