poj3009 Curling 2.0 (good title DFS)

https://vjudge.net/problem/POJ-3009

Finish this question, feel oneself to the DFS understanding should be profound again.

1. In general, the minimum number of steps are used BFS, but this problem because the status record is very troublesome, so you can use DFS.

2. In the use of DFS, MP when a global variable, for the equality of the way, every way after the end of the state must be restored!!! (i.e. code 36-38 lines) Otherwise it will have an effect on other methods.

1#include <iostream>2#include <cstdio>3#include <queue>4#include <cstring>5#include <algorithm>6#include <cmath>7#include <stack>8 #defineLson L, M, rt<<19 #defineRson m+1, R, rt<<1|1Ten #defineIO Ios::sync_with_stdio (false); Cin.tie (0); One #defineINF 0x3f3f3f3f Atypedef unsignedLong Longll; - using namespacestd; - intN, M, mp[ -][ -], si, SJ, GI, GJ, CNT, mini; the intdir[4][2] = {0,1,0, -1,1,0, -1,0}; - voidDfsintXintYintAintBintKintc) - { -     if(k >Ten|| K > Mini)return ; +     if(x = = Gi&&y = =GJ) { -Mini =min (mini, k); +         return ; A     } at     if(A! =0|| B! =0){//not just the beginning . -         if(x<0|| x>=n| | y<0|| y>=m) { -             return;//failed -         } -         Else if(Mp[x][y] = =1){ -             if(c <=1)return ; inc =0; -Mp[x][y] =0;//the resistance is eliminated. toX-=A; +Y-= b;//go back to the previous state -              for(inti =0; I <4; i++){ theDFS (x+dir[i][0], y+dir[i][1], dir[i][0], dir[i][1], K +1, c+1); *             } $X + =A;Panax NotoginsengY + =b; -Mp[x][y] =1;//remember to State recovery!!  the         } +         Else{ ADFS (X+a, y+b, A, B, K, c+1); the         } +     } -     Else{  $          for(inti =0; I <4; i++){ $DFS (x+dir[i][0], y+dir[i][1], dir[i][0], dir[i][1], K +1, c+1); -         }  -     } the } - intMain ()Wuyi { the      while(Cin >> M >>N, N, m) { -Mini =INF; WuCnt=0; -          for(inti =0; I < n; i++){ About              for(intj =0; J < M; J + +){ $CIN >>Mp[i][j]; -                 if(Mp[i][j] = =2) si=i,sj=J; -                 if(Mp[i][j] = =3) gi=i,gj=J; -             } A         } +DFS (SI, SJ,0,0,0,0); the         if(mini = = INF) cout <<"-1"<<Endl; -         Elsecout << Mini <<Endl; $     } the     return 0; the}

poj3009 Curling 2.0 (good title DFS)