Example explanation A *

Topic Links:

http://poj.org/problem?id=2243

The title is simple, a 8*8 chessboard, given the coordinates of two points, ask the knight in chess the minimum number of steps required to go from one point to another. The way to go is similar to Chinese chess inside the horse, eight directions.

Generally will be directly using BFS Search, 8*8 Direct search will not explode, now look at the use of a *.

The key to A * is how to choose a suitable h (x), which is to guarantee: H (x) <= D (x, y) + H (Y), as for the reason, see: http://www.cnblogs.com/be-saber/p/4780564.html

For a step of the way, it is a "day" word.

At this point we can choose Euclidean distance, we can find that H (x), D (x, y), H (Y) is the triangular three-side (possibly three points collinear), then the above conditions are certainly satisfied.

The code is as follows:

1#include <cstdio>2#include <iostream>3#include <queue>4#include <cmath>5#include <algorithm>6#include <cstring>7 using namespacestd;8 9 #definePOS Pair<int, int>Ten Const intMAXM =Ten ; One //h (x) take Euclidean distance A POS s, t; - BOOLVIS[MAXM][MAXM]; - floatGX[MAXM][MAXM]; the intSTEP[MAXM][MAXM]; - intdir[8][2] = {{1,2}, {1, -2}, {-1,2}, {-1, -2}, {2,1}, {2, -1}, {-2,1}, {-2, -1}} ; - // - BOOLjudge (POS x) { +     if(X.first <1|| X.first >8|| X.second <1|| X.second >8)return false ; -     return true ; + } A // at floatCaldis (pos x, pos y) { -     returnsqrt ((x.first-y.first) * (X.first-y.first) + (x.second-y.second) * (x.second-y.second)); - } - // - intsolve () { -memset (Vis,false,sizeof(Vis)); inmemset (GX,0,sizeof(GX)); -memset (step,0,sizeof(step)); topriority_queue< pair<float, Pos> >Heap; +      while( !heap.empty ()) Heap.pop (); -Heap.push (Make_pair (-Caldis (S, t), s)) ; theGx[s.first][s.second] =0 ; *      while( !Heap.empty ()) { $         intval =Heap.top (). First;Panax NotoginsengPOS x =Heap.top (). Second; - Heap.pop (); theVis[x.first][x.second] =true ; +         if(X.first = = T.first && X.second = =T.second) { A             returnStep[x.first][x.second]; the         } +          for(inti =0; I <8; i++ ){ - POS u; $U.first = X.first + dir[i][0], U.second = X.second + dir[i][1] ; $             if(Judge (U) &&!Vis[u.first][u.second]) { -Gx[u.first][u.second] = Gx[x.first][x.second] + sqrt (5) ; -Heap.push (Make_pair (-gx[u.first][u.second]-Caldis (U, T), u)) ; theStep[u.first][u.second] = Step[x.first][x.second] +1 ; -             }Wuyi         } the     }  -      return 0 ; Wu } - // About intMain () { $     ////////freopen ("1234.txt", "R", stdin); -     CharA, B, C, D; -      while(SCANF ("%c%c%c%c\n", &a, &b, &c, &d)! =EOF) { -s = Make_pair (A-'a'+1, B-'0') ; At = Make_pair (C-'a'+1, d'0') ; +         intStep =solve (); theprintf"To get from%c%c to%c%c takes%d Knight moves.\n", A, B, C, D, step); -     } $     return 0 ; the}

Example explanation A *