HDU---2364 Escape【广度优先搜索】

Problem Description

You find yourself trapped in a large rectangular room, made up of large square tiles; some are accessible, others are blocked by obstacles or walls. With a single step, you can move from one tile to another tile if it is horizontally or vertically adjacent (i.e. you cannot move diagonally).
To shake off any people following you, you do not want to move in a straight line. In fact, you want to take a turn at every opportunity, never moving in any single direction longer than strictly necessary. This means that if, for example, you enter a tile from the south, you will turn either left or right, leaving to the west or the east. Only if both directions are blocked, will you move on straight ahead. You never turn around and go back!
Given a map of the room and your starting location, figure out how long it will take you to escape (that is: reach the edge of the room).

 

Input

On the first line an integer t (1 <= t <= 100): the number of test cases. Then for each test case:

a line with two integers separated by a space, h and w (1 <= h, w <= 80), the height and width of the room;

then h lines, each containing w characters, describing the room. Each character is one of . (period; an accessible space), # (a blocked space) or @ (your starting location).
There will be exactly one @ character in each room description.

 

Output

For each test case:

A line with an integer: the minimal number of steps necessary to reach the edge of the room, or -1 if no escape is possible.

 

Sample Input

29 13##############@..........######.#.#.#.##...........##.#.#.#.#.#.##.#.......#.##.#.#.#.#.#.##...........######.#######4 6#.#####.#.###...@#######

 

Sample Output

31-1
SOURCE:点击打开链接
题意：
    自己被困在迷宫中，要求可以前进，左转，右转。然而在遇到能转弯的地方时必须转弯，除非不能转弯或者是转弯才能直行，问至少经过多少步才能走出迷宫。
解析：
   略微变形的BFS，重点在于向左右方向移动。所以用vis数组来记录所处的状态。
代码：
#include <iostream>#include <cstring>#include <queue>using namespace std;int h,w;int vis[85][85][4];    //用来记录所处的位置及状态int dx[4]= {0,1,0,-1};int dy[4]= {1,0,-1,0};char str[85][85];int bfs(const int x,const int y);class node{public:    int x;    int y;    int step;    int face;    node() {};    node (int x,int y,int step,int face);};node::node(int x,int y,int step,int face){    this->x=x;    this->y=y;    this->step=step;    this->face=face;}int main(void){    int t,x,y;    cin>>t;    while(t--)    {        memset(vis,0,sizeof(vis));        cin>>h>>w;        for(int i=0; i<h; i++)            for(int j=0; j<w; j++)            {                cin>>str[i][j];                if(str[i][j]=='@')                {                    x=i;                    y=j;                }            }        cout<<bfs(x,y)<<endl;    }    return 0;}int bfs(const int x,const int y){    queue<node> q;    node n(x,y,0,-1);   //在起始位置可以面向任意方向，所以face为-1    int step,face,nx,ny;    bool flag;   //用来判断是否只能直行    vis[x][y][0]=1,vis[x][y][1]=1;   //在起始位置的四种状态都为0，因为在行走的过程中不能经过起始状态    vis[x][y][2]=1,vis[x][y][3]=1;    q.push(n);    while(q.size())    {        flag=true;        n=q.front();        q.pop();        if(n.x==0||n.x==h-1||n.y==0||n.y==w-1)            return n.step;        for(int i=0; i<4; i++)  //看看能不能转弯        {            if(i%2!=n.face%2)  //只能向左右方向走            {                nx=n.x+dx[i];                ny=n.y+dy[i];                step=n.step+1;                face=i;                if(nx>=0&&nx<h&&ny>=0&&ny<w&&str[nx][ny]=='.')                {                    flag=false;  //只要左右方向能走就走，不管走没走过                    if(!vis[nx][ny][face])  //若没有走过，将其加入队列                    {                        if(nx==0||nx==h-1||ny==0||ny==w-1)                            return step;                        q.push(node(nx,ny,step,face));                        vis[nx][ny][face]=1;                    }                }            }        }        if(flag)  //若不能转弯，则只能直行        {            nx=n.x+dx[n.face];            ny=n.y+dy[n.face];            step=n.step+1;            face=n.face;            if(nx>=0&&nx<h&&ny>=0&&ny<w&&str[nx][ny]=='.'&&!vis[nx][ny][face])            {                if(nx==0||nx==h-1||ny==0||ny==w-1)                    return step;                q.push(node(nx,ny,step,face));                vis[nx][ny][face]=1;            }        }    }    return -1;}