Ural 1145. Rope in the Labyrinth

1145. Rope in the Labyrinthtime limit:0.5 second
Memory limit:64 MBA labyrinth with rectangular form and size mX Nis divided to square cells with sides ' length 1 by lines which is parallel with the labyrinth ' s sides. Each cell of the grid is either occupied or free. It is possible to move from one free cell to another free cells, share a common side with the cell. One cannot move beyond the labyrinth ' s borders. The Labyrinth is designed pretty specially:for any of the cells there is the one-from-one-to-move from one cell to the other. There is a hooks at each cell ' s center. In the labyrinth there is both special free cells, such so if you can connect the hooks of the those of both cells with a rope, The Labyrinth ' s secret door'll be automatically opened. The problem is to prepare a shortest rope so can guarantee, you always can connect the hooks of the those of both cells with the Prepared rope regardless their position in the labyrinth. Inputthe First line contains integers Nand m(3≤ N, m≤820). The next lines describe the labyrinth. Each of the next mLines contains NCharacters. Each character are either "#" or ".", with "#" indicating a occupied cell, and "." indicating a free cell. Outputprint out of the "measured in the number of cells" of the required rope. Sample

input	Output
7 6########.#.####.#.####.#.#.##.....########	8

Tags:Graph theory()difficulty:425 Test Instructions: It is possible to walk, #不可以走. All the. Make up a tree and ask the diameter of the tree. Analysis: Finding the diameter of the tree. BFS can be.

1 /**2 Create by Yzx-stupidboy3 */4#include <cstdio>5#include <cstring>6#include <cstdlib>7#include <cmath>8#include <deque>9#include <vector>Ten#include <queue> One#include <iostream> A#include <algorithm> -#include <map> -#include <Set> the#include <ctime> -#include <iomanip> - using namespacestd; -typedefLong LongLL; +typedefDoubleDB; - #defineMIT (2147483647) + #defineINF (1000000001) A #defineMLL (1000000000000000001LL) at #defineSZ (x) ((int) (x). Size ()) - #defineCLR (x, y) memset (x, y, sizeof (x)) - #definePUF Push_front - #definePub push_back - #definePOF Pop_front - #definePOB pop_back in #defineFT first - #defineSD Second to #defineMk Make_pair +  -InlineintGetint () the { *     intRet =0; $     CharCh =' ';Panax Notoginseng     BOOLFlag =0; -      while(! (Ch >='0'&& Ch <='9')) the     { +         if(Ch = ='-') Flag ^=1; ACh =GetChar (); the     } +      while(Ch >='0'&& Ch <='9') -     { $RET = RET *Ten+ Ch-'0'; $Ch =GetChar (); -     } -     returnFlag? -Ret:ret; the } - Wuyi Const intN =1010; the Const intDx[] = {-1,0,1,0}, dy[] = {0, -1,0,1}; - intN, M; Wu CharGraph[n][n]; - intDp[n][n]; Aboutqueue<pair<int,int> >que; $  -InlinevoidInput () - { -scanf"%d%d", &m, &n); A      for(inti =0; I < n; i++) scanf ("%s", Graph[i]); + } the  -InlineBOOLCheck (intXinty) $ { the     if(X <0|| Y <0|| X >= N | | Y >= m)return 0; the     if(Graph[x][y]! ='.')return 0; the     return 1; the } -  inInlinevoidBfs (intSxintSy) the { the      for(inti =0; I < n; i++) About          for(intj =0; J < M; J + +) theDP[I][J] =INF; the Que.push (MK (SX, SY)); theDp[sx][sy] =0; +      while(sz (que)) -     { the         intUX = Que.front (). ft, Uy =Que.front (). SD;Bayi Que.pop (); the          for(intt =0; T <4; t++) the         { -             intVX = UX + dx[t], vy = Uy +Dy[t]; -             if(Check (VX, VY) && Dp[vx][vy] > Dp[ux][uy] +1) the             { theDp[vx][vy] = Dp[ux][uy] +1; the Que.push (MK (VX, VY)); the             } -         } the     } the } the 94InlinevoidGetmax (int&AMP;PX,int&py) the { the     intMX =-INF; the      for(inti =0; I < n; i++)98          for(intj =0; J < M; J + +) About             if(MX < dp[i][j] && Dp[i][j] <INF) -             {101MX =Dp[i][j];102px = i, py =J;103             }104 } the 106InlinevoidSolve ()107 {108     BOOLFlag =0;109      for(inti =0; I < n &&!flag; i++) the          for(intj =0; J < m &&!flag; J + +)111             if(Graph[i][j] = ='.') the             {113 Bfs (i, j); theFlag =1; the             } the 117     intpx, py;118 getmax (px, py);119 Bfs (px, py); - 121 getmax (px, py);122printf"%d\n", Dp[px][py]);123 }124  the intMain ()126 {127Freopen ("a.in","R", stdin); - Input ();129 Solve (); the     return 0;131}

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Ural 1145. Rope in the Labyrinth