Codeforces Round #354 (Div. 2) D.Theseus and labyrinth（BFS）

题目链接：http://codeforces.com/contest/676/problem/D

D. Theseus and labyrinth

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Theseus has just arrived to Crete to fight Minotaur. He found a labyrinth that has a form of a rectangular field of size n × m and
consists of blocks of size 1 × 1.

Each block of the labyrinth has a button that rotates all blocks 90 degrees
clockwise. Each block rotates around its center and doesn't change its position in the labyrinth. Also, each block has some number of doors (possibly none). In one minute, Theseus can either push the button in order to rotate all the blocks 90 degrees
clockwise or pass to the neighbouring block. Theseus can go from block A to some neighbouring block B only
if block A has a door that leads to block B and
block B has a door that leads to block A.

Theseus found an entrance to labyrinth and is now located in block (xT, yT) —
the block in the row xT and
column yT.
Theseus know that the Minotaur is hiding in block (xM, yM) and
wants to know the minimum number of minutes required to get there.

Theseus is a hero, not a programmer, so he asks you to help him.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 1000) —
the number of rows and the number of columns in labyrinth, respectively.

Each of the following n lines contains m characters,
describing the blocks of the labyrinth. The possible characters are:

«+» means this block has 4 doors (one door
to each neighbouring block);

«-» means this block has 2 doors — to the
left and to the right neighbours;

«|» means this block has 2 doors — to the
top and to the
4000
bottom neighbours;

«^» means this block has 1 door — to the
top neighbour;

«>» means this block has 1 door — to the
right neighbour;

«<» means this block has 1 door — to the
left neighbour;

«v» means this block has 1 door — to the
bottom neighbour;

«L» means this block has 3 doors — to all
neighbours except left one;

«R» means this block has 3 doors — to all
neighbours except right one;

«U» means this block has 3 doors — to all
neighbours except top one;

«D» means this block has 3 doors — to all
neighbours except bottom one;

«*» means this block is a wall and has no doors.

Left, right, top and bottom are defined from representing labyrinth as a table, where rows are numbered from 1 to n from
top to bottom and columns are numbered from 1 to m from
left to right.

Next line contains two integers — coordinates of the block (xT, yT) (1 ≤ xT ≤ n, 1 ≤ yT ≤ m),
where Theseus is initially located.

Last line contains two integers — coordinates of the block (xM, yM) (1 ≤ xM ≤ n, 1 ≤ yM ≤ m),
where Minotaur hides.

It's guaranteed that both the block where Theseus starts and the block where Minotaur is hiding have at least one door. Theseus and Minotaur may be initially located at the same block.

Output

If Theseus is not able to get to Minotaur, then print -1 in the only line of the output. Otherwise, print the minimum number of minutes required to get to the
block where Minotaur is hiding.

Examples

input

2 2
+*
*U
1 1
2 2

output

-1

input

2 3
<><
><>
1 1
2 1

output

4

Note

Assume that Theseus starts at the block (xT, yT) at
the moment 0.

题目大意：一个n*m的迷宫，然后迷宫每一格都有符号，符号意思如题，太多了，然后可以花费一秒钟使每个符号顺时针旋转九十度如果是‘L’则变成'U'

如果能走的路必须是互通的，比如a能到b，那么b也必须要能到a，然后给出初始位置x,y然后要到的终点，如果不能到达输出-1，能到达输出最小步数，

直接上bfs，不会超时的

#include <bits/stdc++.h>
using namespace std;
char mp[4][1010][1010];
int vis[1010][1010][4];
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int n,m,ans;
int x,y,x2,y2;
struct node
{
    int x,y,s,t;
};
node st;
char change(char x) //顺时针旋转后的结果
{
    if(x=='+') 
        return '+';
    if(x=='-') 
        return '|';
    if(x=='|') 
        return '-';
    if(x=='^') 
        return '>';
    if(x=='>') 
        return 'v';
    if(x=='v') 
        return '<';
    if(x=='<') 
        return '^';
    if(x=='L') 
        return 'U';
    if(x=='R') 
        return 'D';
    if(x=='U') 
        return 'R';
    if(x=='D') 
        return 'L';
    if(x=='*') 
        return '*';
}
int YR(char x)  //能不能往右
{
    if(x=='+'||x=='-'||x=='>'||x=='L'||x=='U'||x=='D') 
        return 1;
    return 0;
}
int YL(char x)  //能不能往左
{
    if(x=='+'||x=='-'||x=='<'||x=='R'||x=='U'||x=='D') 
        return 1;
    return 0;
}
int YU(char x)  //能不能向上
{
    if(x=='+'||x=='|'||x=='^'||x=='L'||x=='R'||x=='D') 
        return 1;
    return 0;
}
int YD(char x)  //能不能向下
{
    if(x=='+'||x=='|'||x=='v'||x=='L'||x=='U'||x=='R') 
        return 1;
    return 0;
}
int check(node x,node y,int k)  //判断能不能互通
{
    int t = x.s;
    char a = mp[t][x.x][x.y],b = mp[t][y.x][y.y];
    if(k == 2)
    {
        if(YR(a) && YL(b)) 
            return 1;
        return 0;
    }
    if(k==3)
    {
        if(YR(b) && YL(a)) 
            return 1;
        return 0;
    }
    if(k==0)
    {
        if(YD(a) && YU(b)) 
            return 1;
        return 0;
    }
    if(k==1)
    {
        if(YU(a)&&YD(b)) 
            return 1;
        return 0;
    }
    return 0;
}
void bfs()      //搜索
{
    queue<node>q;
    q.push(st);
    vis[x][y][0] = 1;
    while(!q.empty())
    {
        node now=q.front();
        q.pop();
        if(now.x == x2 && now.y == y2)  //到达终点
        {
            ans = now.t;
            break;
        }
        if(!vis[now.x][now.y][(now.s+1)%4])  //旋转
        {
            node next = now;
            next.s = (now.s+1)%4;
            next.t += 1;
            q.push(next);
            vis[next.x][next.y][next.s] = 1;
        }
        for(int i = 0 ; i < 4 ; i++)  //四个方向
        {
            node next = now;
            next.x += dx[i];
            next.y += dy[i];
            next.t += 1;
            if(next.x < 0 || next.x >= n) 
                continue;
            if(next.y < 0 ||next.y >= m) 
                continue;
            if(vis[next.x][next.y][next.s]) 
                continue;
            if(!check(now,next,i)) 
                continue;
            q.push(next);
            vis[next.x][next.y][next.s] = 1;
        }
    }
    return ;
}
int main()
{
    ans=-1;
    while (~scanf("%d %d",&n,&m))
    {
        for(int i = 0 ; i < n ; i++) 
            scanf("%s",mp[0][i]);
        for(int l = 1 ; l < 4 ; l++)
        {
            for(int i = 0 ; i < n ; i++)
            {
                for(int j = 0 ; j < m ; j++)
                {
                    mp[l][i][j] = change(mp[l-1][i][j]);
                }
            }
        }
        scanf("%d%d%d%d",&x,&y,&x2,&y2);
        x--,y--,x2--,y2--;
        st.x = x,st.y = y,st.s = 0,st.t = 0;
        bfs();
        printf("%d\n",ans);
    }
   
    return 0;
}