Hdu 5335 Walk out (bfs+ looking for path)

problem Description

In an n∗m maze, the Right-bottom corner isThe exit (position (N,M) isThe exit). In every position of ThisMaze, there isEither a0or a1written on it. An explorer gets lostinch  ThisGrid. His position now is(1,1), and he wants to go to the exit. Since to arrive at the exit isEasy forHim, he wants to DoSomething more difficult. At first, he'll write down the number on position. Every time, he could make a move to one adjacent position (both positions is adjacent if and only if they share an edge). While walking, he would write down the number on the position he's on to the end of his number. When finished, he'llGetA binary number. Please determine the minimum value of ThisNumberinchBinary system.

Input

 is a single integer T (t=), indicating the number of testcases. For each testcase, the first line contains the integers n and m (1≤n,m≤string
     of length m, which represents i-th row of the maze.

Output

inch 0  is 0 (in this case0 instead).

Sample Input

2 2 2  One  One 3 3 001 111 101

Sample Output

111 101

Author XJZX

Source multi-university Training Contest 4

1, if mp[0][0]==0, first BFS once, the leading 0 is removed. BFs got Wx,wy.

2, then is to find the smallest path. A walking count for (int i=wx+wy;i<n+m-2;i++), which enumerates all possible (x, y) x+y==step based on this number of steps, and if 0 is encountered, fetch 0, always found (n-1,m-1)

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <queue>5#include <stdlib.h>6 using namespacestd;7 #defineN 10068 intn,m;9 intMp[n][n];Ten intVis[n][n]; One structnode{ A     intx, y; - }st; - intdirx[]={1,0,-1,0}; the intdiry[]={0,1,0,-1}; - intWx,wy; - voidBFs () { -Queue<node>Q; + Q.push (ST); - Node t1,t2; +      while(!Q.empty ()) { At1=Q.front (); at Q.pop (); -          for(intI=0;i<4; i++){ -t2.x=t1.x+Dirx[i]; -t2.y=t1.y+Diry[i]; -             if(Vis[t2.x][t2.y])Continue; -             if(t2.x<0|| T2.x>=n | | t2.y<0|| T2.Y&GT;=M)Continue; invis[t2.x][t2.y]=1; -             if(mp[t2.x][t2.y]==0){ to Q.push (T2); +             } -             if(wx+wy<t2.x+t2.y) { thewx=t2.x; *wy=t2.y; $             }Panax Notoginseng         } -     } the } + intMain () A { the     intT; +scanf"%d",&t); -      while(t--){ $scanf"%d%d",&n,&m); $         Chars[1006]; -          for(intI=0; i<n;i++){ -scanf"%s", s); the              for(intj=0; j<m;j++){ -mp[i][j]=s[j]-'0';Wuyi             } the         } -st.x=0; Wust.y=0; -memset (Vis,0,sizeof(Vis)); Aboutvis[0][0]=1; $wx=0; -wy=0; -         if(mp[0][0]==0){ - BFS (); A         } +         if(mp[wx][wy]==0){ theprintf"0\n"); -             Continue; $         } theprintf"1"); the         intnowflag=0; the          for(inti=wx+wy;i<n+m-2; i++){ the             intflag=0; -             intstep=i; in              for(intj=0; j<=step;j++){ the                 intx=J; the                 inty=i-J; About                 if(x<0|| X>=n | | y<0|| Y&GT;=M)Continue; the                 if(Nowflag && Mp[x][y])Continue; the                 if(!vis[x][y])Continue; the                  +                  for(intk=0;k<2; k++){ -                     intsx=x+Dirx[k]; the                     intsy=y+Diry[k];Bayi                     if(sx<0|| Sx>=n | | sy<0|| SY&GT;=M)Continue; thevis[sx][sy]=1; the                     if(mp[sx][sy]==0){ -flag=1; -                     } the                 } the                  the             } thenowflag=Flag; -             if(flag) { theprintf"0"); the             } the             Else{94printf"1"); the             } the         } theprintf"\ n");98     } About     return 0; -}

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Hdu 5335 Walk out (bfs+ looking for path)