POJ 3050 Map 5-bit number problem DFS algorithm

Test instructions: On a 5*5 map, jump five times from any position up or down, forming a number. Q: How many non-repeating numbers are there?

Idea: DFS

1. Jump 5 times from any location, indicating that each location requires traversal.
2. form a number: Number*10+map[dx][dy]
3. A number that is not duplicated, stored with set (set)
4. Only need to jump each time the number of steps plus 1, and can jump position, as long as no more than the range can be, that a position can be repeated jump

Code to solve the problem:

#include <iostream>#include<cstdio>#include<Set>using namespacestd;intmap[5][5];Set<int>results;Const intdir[4][2]{    { 0,1},{0,-1},{1,0},{ -1,0 }};voidDfsConst int& X,Const int& Y,Const int& Step,Const int&Number ) {    if(Step = =5) {Results.insert (number); return; }     for(inti =0; I <4; i++)    {        intDX = x + dir[i][0]; intDy = y + dir[i][1]; if(DX >=0&& DX <5&& Dy >=0&& Dy <5) Dfs (dx, dy, step+1, number *Ten+Map[dx][dy]); }}intMain () { for(inti =0; I <5; i++)         for(intj =0; J <5; J + +) scanf ("%d", &Map[i][j]);  for(inti =0; I <5; i++)         for(intj =0; J <5; J + +) Dfs (i, J,0, Map[i][j]); printf ("%d\n", Results.size ()); return 0;}

POJ 3050 Map 5-bit number problem DFS algorithm