poj3984迷宫问题(bfs打印路径)

题目链接:http://poj.org/problem?id=3984

比较简单,我是用string记录每次走的方向,最后输出。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<queue>
 4 #include<string>
 5 #include<cstring>
 6 using namespace std;
 7 int dir[4][2]={0,1,0,-1,1,0,-1,0};
 8 int vis[5][5];
 9 int pic[5][5];
10 struct node
11 {
12     int x,y;
13     int parx,pary;
14     string s;
15 
16 };
17 node now,nex;
18 int read()
19 {
20     for(int i=0;i<5;i++)
21         for(int j=0;j<5;j++)
22         if(scanf("%d",&pic[i][j])==EOF) return 0;
23         return 1;
24 }
25 string bfs()
26 {
27    queue<node>q;
28    now.x=0;
29    now.y=0;
30    now.s="";
31    vis[0][0]=1;
32    q.push(now);
33    while(!q.empty())
34    {
35        now=q.front();
36        q.pop();
37        if(now.x==4&&now.y==4) return now.s;
38        for(int i=0;i<4;i++)
39        {
40            nex.x=now.x+dir[i][0];
41            nex.y=now.y+dir[i][1];
42            if(nex.x>=0&&nex.x<5&&nex.y>=0&&nex.y<5&&!vis[nex.x][nex.y]&&pic[nex.x][nex.y]==0)
43            {
44                vis[nex.x][nex.y]=1;
45                switch(i)
46                {
47                case 0:nex.s=now.s+'0';break;
48                case 1:nex.s=now.s+'1';break;
49                case 2:nex.s=now.s+'2';break;
50                case 3:nex.s=now.s+'3';break;
51                }
52                q.push(nex);
53            }
54        }
55    }
56 }
57 void print(string s)  //打印路径
58 {
59 
60   int len=s.length();
61       int x=0,y=0;
62       printf("(%d, %d)
",x,y);
63   for(int i=0;i<len;i++)
64   {
65       x=x+dir[s[i]-'0'][0];
66       y=y+dir[s[i]-'0'][1];
67     printf("(%d, %d)
",x,y);
68   }
69 
70 }
71 int main()
72 {
73     while(read())
74     {
75         memset(vis,0,sizeof(vis));
76         print(bfs());
77     }
78 }
原文地址:https://www.cnblogs.com/yijiull/p/6613249.html