Easy Graph problem? (A simple graph thesis?) BFS + Priority Queue

Easy Graph problem?

Topic Abstraction: Give you a matrix of n * M. Each lattice is a number or an obstacle ' * '.  First question: The shortest path from the starting point to the end point. Second question: Require the same direction can not walk two times, that is, each to arrive at a lattice, need to change direction. Ask the shortest circuit. But there is no output-1.

Analysis: Can be directly BFS search. The first question has a minv[ms][ms] array record the shortest distance to reach (i,j). Second asked with Flag[ms][ms][4] and record (i,j) lattice K direction whether to walk.

Due to the 2<=n,m<=500, the range is larger. So priority queue optimization is required. Otherwise time out.

1#include <cstdio>2#include <algorithm>3#include <queue>4 using namespacestd;5typedefLong LongLL;6 Const intINF =0x5fffff;7 Const intMS =505;8 9 structNode {Ten     intx, Y, dir, cost; One     BOOL operator< (ConstNode & A)Const { A         returnCost >A.cost; -     } - }; the intPic[ms][ms]; - intdir[4][2] = {0,1,1,0,0,-1,-1,0}; - intflag[ms][ms][4]; - intMinv[ms][ms]; +  - intN, M, R1, C1, R2, C2; + intKase =1; A node s,t; at  - voidbfs1 () { -MINV[R1][C1] =PIC[R1][C1]; -S.x =R1; -S.y =C1; -S.cost =PIC[R1][C1]; inPriority_queue<node>que; - Que.push (s); to      while(!Que.empty ()) { +s =que.top (); - Que.pop (); the          for(inti =0; I <4; i++) { *             intx = s.x + dir[i][0]; $             inty = s.y + dir[i][1];Panax Notoginseng             if(X >0&& x <= n && y >0&& y <= M &&Pic[x][y]) { -T.x =x; theT.y =y; +T.cost = S.cost +Pic[x][y]; A                 if(x = = R2 && y = =C2) { theprintf"%d", t.cost); +                     return ; -                 } $                 if(T.cost <Minv[x][y]) { $Minv[x][y] =T.cost; - Que.push (t); -                 } the             } -         }Wuyi     } theprintf"-1"); - } Wu  - voidbfs2 () { AboutS.x =R1; $S.y =C1; -S.cost =PIC[R1][C1]; -Priority_queue<node>que; -      for(inti =0; I <4; i++) { AFlag[r1][c1][i] =1; +         intx = s.x + dir[i][0]; the         inty = s.y + dir[i][1]; -         if(X >0&& x <= n && y >0&& y <= M &&Pic[x][y]) { $Flag[x][y][i] =1; theT.x =x; theT.y =y; theT.cost = S.cost +Pic[x][y]; theT.dir =i; - Que.push (t); in         } the     } the      while(!Que.empty ()) { Abouts =que.top (); the Que.pop (); the          for(inti =0; I <4; i++) { the             if(i = =S.dir) +                 Continue; -T.x = s.x + dir[i][0]; theT.y = S.y + dir[i][1];Bayi             if(T.x >0&& t.x <= n && t.y >0&& T.y <= m &&Pic[t.x][t.y]) { theT.cost = S.cost +Pic[t.x][t.y]; theT.dir =i; -                 if(T.x = = R2 && T.y = =C2) { -printf"%d", t.cost); the                     return ; the                 } the                 if(Flag[t.x][t.y][t.dir] = =0) { theFlag[t.x][t.y][t.dir] =1; - Que.push (t); the                 } the             } the 94         } the     } theprintf"-1"); the }98  About intMain () { -     CharStr[ms];101      while(SCANF ("%d%d%d%d%d%d", &n, &m, &r1, &c1, &AMP;R2, &c2)! =EOF) {102          for(inti =1; I <= N; i++)103              for(intj =1; J <= M; J + +) {104scanf"%s", str); the                 if(str[0] =='*') {106PIC[I][J] =0;107                 }108                 Else {109SSCANF (str,"%d", &pic[i][j]);//initialization Diagram the                 }111MINV[I][J] = INF;//(I,J) Minimum cost initialization the                  for(intK =0; K <4; k++)113FLAG[I][J][K] =0;//(I,J) four-direction initialization the             } theprintf"Case %d:", kase++); the BFS1 ();117 BFS2 ();118printf"\ n");119     } -     return 0;121}

Easy Graph problem? (A simple graph thesis?) BFS + Priority Queue