Sicily Maze (BFS Compute Maze Shortest path)

	Maze.

	Time Limit:1sec Memory limit:256mbdescription Master Tang is kidnapped by a monster and put in a maze, and Mr. Sha wants to rescue him. The maze is an n*m matrix. There is types rooms in the maze, 1 for safe and 0 for dangerous. Of course, Mr Sha can only be stay in the safe. Mr. Sha is today in the top left corner. The lower right corner guest (N,M) Mr. Sha can move to another safe, and down, left or right direction. Each move takes 1 unit of time. Mr. Sha wants to find Master Tang as soon as possible. Can-You-help Mr. Sha to count the least time-to-get to the "where Master Tang is?" Input The first line contains an integer t (0<t<=10), which means t test cases followed. The first line of all test case contains 2 integers n (1≤n≤120) and M (1≤n≤120), which is the size of the maze. The next n lines input the maze, each line contains m number (0 or 1) seperated by one space. The next line contains 2 integers x (1≤n≤120) and Y (1≤n≤120), which means Master Tang are in the (x, y). Output For each test case, the output of the answer in a line:the least time for Mr. Sha to get to Master Tang. If Mr. Sha can ' t get to Mast Tang, output-1. Sample Inputcopy sample input to Clipboard 22 21 00 12 24 4 1 1 1 01 0 1 11 0 1 11 1 1 03 4 Sample Output -15 Title: Starting from (a), go to the end of the input to save people. In the middle of the room, 1 is safe, 0 means danger. If it can be reached, go to the shortest path. cannot be output-1. This is a maze of the topic, simple calculation of the shortest path can be. But stuck inside the queue. Lead to wasted time. You can do it with a vector. 1#include <iostream>2#include <cmath>3#include <queue>4#include <vector>5 6 using namespacestd;7 8 structnode9 {Ten intx; One inty; A }; - - //Four Directions the intmove[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; - intn,m,room[1001][1001],x,y; - - //is this the way to go ? + BOOLSafeintXinty) - { + if(X >0&& y >0&& x <= n && y <= m && room[x][y] = =1) A { at return true; - } - return false; - } - - intMain () in { - intT; to +Cin>>T; - the while(t--) * { $Cin>>n>>m;Panax NotoginsengVector <node> v[1001]; - node No; the + for(inti =1; I <= N; i++) A { the for(intj =1; J <= M; J + +) + { -Cin>>Room[i][j]; $ } $ } -Cin>>x>>y; -no.x =1; theNO.Y =1; -v[0].push_back (no);Wuyi intK =0; the intFlag =0; - //use vector to do breadth First search Wu while(1) - { About if(V[k].begin () = =v[k].end ()) $ { - Break; - } -Vector <node>:: iterator it; A for(It=v[k].begin (); It!=v[k].end (); it++) + { theNode F = *it; - for(inti =0; I <4; i++) $ { the node tmp; thetmp.x = f.x + move[i][0]; theTMP.Y = F.y + move[i][1]; the if(tmp.x = = x && tmp.y = =y) - { inFlag =1; the Break; the } About if(Safe (tmp.x,tmp.y)) the { theV[k +1].push_back (TMP); theROOM[TMP.X][TMP.Y] =0; + } - } the }Bayik++; the if(Flag = =1) the { - Break; - } the } the if(flag) the { thecout<<k<<Endl; - } the Else the { thecout<<"-1"<<Endl;94 } the the } the return 0;98}

Sicily Maze (BFS Compute Maze Shortest path)