Poj3984 迷宫问题
![]()
1 #include <iostream>
2 #include <algorithm>
3 #include <cstdio>
4 #include <queue>
5 using namespace std;
6 int maze[5][5];
7 int dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};
8
9 //用结构体来记录更方便
10 struct Node
11 {
12 int x ,y ;
13 int prex ,prey; //前一节点坐标
14 int dis ; //记录是第几层访问的节点
15 } s[5][5];
16
17
18 void bfs(int x, int y)
19 {
20 //队列来实现bfs
21 queue <Node> q;
22 q.push(s[0][0]); //加入头节点
23 s[0][0].dis = 0;
24 s[0][0].x=0;
25 s[0][0].y=0;
26 while(!q.empty())
27 {
28 Node temp = q.front();
29 int tx = temp.x;
30 int ty = temp.y;
31 int tdis = temp.dis;
32
33 if(tx == 4 && ty == 4) //终止条件
34 return;
35
36 for(int i = 0; i < 4; i++)
37 {
38 int row = tx + dir[i][0];
39 int col = ty + dir[i][1];
40 if(row >=0 && row < 5 && col >=0 && col < 5 && maze[row][col] != 1)
41 {
42 maze[row][col] = 1;
43 s[row][col].x = row;
44 s[row][col].y = col;
45 s[row][col].prex = tx;
46 s[row][col].prey = ty;
47 s[row][col].dis = tdis + 1; //有了这一步，便可知道最短路径的长度！
48 q.push(s[row][col]);
49 }
50 }
51 q.pop();
52
53
54
55 }
56
57 }
58
59 //递归打印路径！从后往前，打印从前往后
60 void print_path(int x,int y)
61 {
62
63 if(x == 0 && y == 0) //终止条件，打印第一个
64 {
65 cout<<"("<<s[0][0].x<<", "<<s[0][0].y<<")"<<endl;
66 return;
67 }
68
69 int prex = s[x][y].prex;
70 int prey = s[x][y].prey;
71 print_path(prex,prey);
72 cout<<"("<<s[x][y].x<<", "<<s[x][y].y<<")"<<endl;
73 }
74
75 int main()
76 {
77 for(int i = 0; i < 5; i++)
78 for(int j = 0; j < 5; j++)
79 cin>>maze[i][j];
80 bfs(0,0);
81 //cout<<s[4][4].dis<<endl;
82 print_path(4,4);
83 return 0;
84 }
View Code