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1000. Maze
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1000. Maze
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Description
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 are two types rooms in the maze, 1 for safe and 0 for dangerous. Of course, Mr. Sha can only stay in the safe room. Mr. Sha is now in the top left corner room, the room (1,1). The lower right corner room is room (n,m)
Mr. Sha can move to another safe room in up, 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 room where Master Tang is in?
Input
The first line contains an integer t (0<t<=10), which means t test cases followed. The first line of each 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 is in the room (x,y).
Output
For each test case, output the answer in a single line: the least time for Mr. Sha to get to Master Tang. If Mr. Sha can‘t get to Mast Tang, output -1. Sample Input
2 2 2 1 0 0 1 2 2 4 4 1 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 3 4 Sample Output
-1 5 1 #include<iostream> 2 #include<cmath> 3 #include<queue> 4 #include<vector> 5 6 using namespace std; 7 8 struct node 9 { 10 int x; 11 int y; 12 }; 13 14 //四个方向 15 int move[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; 16 int n,m,room[1001][1001],x,y; 17 18 //这条路是否可走 19 bool safe(int x,int y) 20 { 21 if(x > 0 && y > 0 && x <= n && y <= m && room[x][y] == 1) 22 { 23 return true; 24 } 25 return false; 26 } 27 28 int main() 29 { 30 int T; 31 32 cin>>T; 33 34 while(T--) 35 { 36 cin>>n>>m; 37 vector <node> v[1001]; 38 node no; 39 40 for(int i = 1; i <= n; i++) 41 { 42 for(int j = 1; j <= m; j++) 43 { 44 cin>>room[i][j]; 45 } 46 } 47 cin>>x>>y; 48 no.x = 1; 49 no.y = 1; 50 v[0].push_back(no); 51 int k = 0; 52 int flag = 0; 53 //用vector做广度优先搜索 54 while(1) 55 { 56 if(v[k].begin() == v[k].end()) 57 { 58 break; 59 } 60 vector <node>::iterator it; 61 for(it=v[k].begin();it!=v[k].end();it++) 62 { 63 node f = *it; 64 for(int i = 0; i < 4; i++) 65 { 66 node tmp; 67 tmp.x = f.x + move[i][0]; 68 tmp.y = f.y + move[i][1]; 69 if(tmp.x == x && tmp.y == y) 70 { 71 flag = 1; 72 break; 73 } 74 if(safe(tmp.x,tmp.y)) 75 { 76 v[k + 1].push_back(tmp); 77 room[tmp.x][tmp.y] = 0; 78 } 79 } 80 } 81 k++; 82 if(flag == 1) 83 { 84 break; 85 } 86 } 87 if(flag) 88 { 89 cout<<k<<endl; 90 } 91 else 92 { 93 cout<<"-1"<<endl; 94 } 95 96 } 97 return 0; 98 }
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原文地址：http://www.cnblogs.com/huang22/p/5466916.html
