"Bfs+ priority queue" Poj2312-battle City

Ideas

The topic "can be launched in a straight line to break the brick wall" may confuse many people, in fact, can be interpreted as equivalent to "through the brick wall of time for 2 units", so the problem is solved. The first encounter can not be very good grasp, the second basic can be used as a water problem.

"Error Point"

1. Do not use the bare BFS. The practical idea of wide search is to put the shortest arrival time in the first team, so that the first time to reach the end is the minimum value. The time through the brick wall is two units, which may not be the time minimum after passing through the brick wall. Use a priority queue to solve this problem.

2.c++ Memset is initialized to memset (vis,0,sizeof (VIS)), and many people may write memset (vis,sizeof (Vis), 0).

1#include <iostream>2#include <cstdio>3#include <iostream>4#include <queue>5 using namespacestd;6 Const intmaxn= -+5;7 structRec8 {9     intX,y,cost;Ten     BOOL operator< (ConstRec &x)Const One     { A         return(Cost >x.cost); -     } - }; the CharMAP[MAXN][MAXN]; - intM,n,yx,yy; -  - voidInit () + { - GetChar (); +      for(intI=0; i<m;i++) A     { at          for(intj=0; j<n;j++) -         { -scanf"%c",&map[i][j]); -             if(map[i][j]=='Y') -             { -yx=i;yy=J; in             } -         } to GetChar (); +     } - } the  * intBFS () $ {Panax Notoginseng     intVIS[MAXN][MAXN]; -     intdx[4]={0,0,1,-1}; the     intdy[4]={1,-1,0,0}; +Priority_queue<rec>que; Amemset (Vis,0,sizeof(Vis)); thevis[yx][yy]=1; + Rec now; -now.x=yx;now.y=yy;now.cost=0; $ Que.push (now); $      while(!que.empty ()) -     { -Rec head=que.top (); the         if(map[head.x][head.y]=='T')return(head.cost); - Que.pop ();Wuyi          for(intI=0;i<4; i++) the         { -             inttempx=head.x+dx[i],tempy=head.y+Dy[i]; Wu             if(tempx<0|| tempx>=m| | tempy<0|| tempy>=n| | map[tempx][tempy]=='R'|| map[tempx][tempy]=='S'|| Vis[tempx][tempy])Continue; -vis[tempx][tempy]=1; Aboutnow.x=tempx;now.y=Tempy; $now.cost=head.cost+1; -             if(map[tempx][tempy]=='B') now.cost++; - Que.push (now); -         } A     } +     return(-1); the } -  $ intMain () the { the      while(SCANF ("%d%d", &m,&n)! =EOF) the     { the         if(M==n && n==0) Break; - init (); inCOUT&LT;&LT;BFS () <<Endl; the     } the     return 0; About}

"Bfs+ priority queue" Poj2312-battle City