poj 3050 dfs(暴力)

Hopscotch
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4259 Accepted: 2848
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

分析：可以对每个格子做深度优先遍历，构造出所有数字，但要注意不要重复计数。在这里，我使用了map来保存已构造出的数字，结果就是map中的元素个数。

java ac代码

import java.util.HashMap;import java.util.Map;import java.util.Scanner;public class Main {    /**     * @param args     */    static int e[][]=new int[5][5];    static Map<Integer,Integer> map=new HashMap<Integer,Integer>();    static int dx[] = {-1, 1, 0, 0};    static int dy[] = {0, 0, -1, 1};    public static void main(String[] args) {        // TODO Auto-generated method stub        Scanner scan=new Scanner(System.in);        for(int i=0;i<5;i++){            for(int j=0;j<5;j++){                e[i][j]=scan.nextInt();            }        }        for(int i=0;i<5;i++){            for(int j=0;j<5;j++){                dfs(i,j,1,e[i][j]);            }        }        System.out.println(map.size());    }    private static void dfs(int x, int y, int k, int num) {        // TODO Auto-generated method stub        if(k==6){            if(!map.containsKey(num)){//注意判断数字是否重复                map.put(num, 0);                return ;            }            return ;        }        for(int i=0;i<4;i++){            int nx=x+dx[i];            int ny=y+dy[i];            if(nx>=0&&ny>=0&&nx<5&&ny<5){                k++;                dfs(nx,ny,k,num*10+e[nx][ny]);                k--;            }        }    }}