Theseus has just arrived to Crete to fight Minotaur. He found a labyrinth that has a form of a rectangular field of sizen × 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.
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 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 nfrom 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.
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.
2 2
+*
*U
1 1
2 2
-1
2 3
<><
><>
1 1
2 1
4
Assume that Theseus starts at the block (xT, yT) at the moment 0.
题意：给定一个图表，然后是开始位置和目标位置，从开始到目标的最少时间，不存在输出-1. 每一秒可以向能走的方向走一步或者顺时针旋转全部按钮90度。注意如果你要从A->B，那么必须B->A才能从A走到B。
思路：BFS，按照思路模拟吧，比较麻烦的一道题目。 开始以为会TLE用的A*发现出来的不一定最优解。改成普通的队列就AC了。
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<bitset>
#include<queue>
#include<vector>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define PI 3.14159
const int MAXN = + ;
char g[MAXN][MAXN];
int dist[][] = { , , , -, , , -, }; //0:right 1:left 2:down 3:up
int n, m, vis[MAXN][MAXN][];
map<char, vector<int> >Move;
struct Node
{
int x, y;
int step;
char str;
int cnt;
int f;
Node(int a = , int b = , int c = , char d = '*', int e = , int ff = ) :x(a), y(b), step(c), str(d), cnt(e), f(ff){};
};
struct cmp
{
bool operator ()(const Node &a, const Node &b)const
{
return a.f>b.f;
}
};
void init() //初始化每个按钮能走的方向
{
vector<int> op1; op1.push_back(); op1.push_back(); op1.push_back(); op1.push_back(); Move['+'] = op1;
vector<int> op2; op2.push_back(); op2.push_back(); Move['-'] = op2;
vector<int> op3; op3.push_back(); op3.push_back(); Move['|'] = op3;
vector<int> op4; op4.push_back(); Move['^'] = op4;
vector<int> op5; op5.push_back(); Move['>'] = op5;
vector<int> op6; op6.push_back(); Move['<'] = op6;
vector<int> op7; op7.push_back(); Move['v'] = op7;
vector<int> op8; op8.push_back(); op8.push_back(); op8.push_back(); Move['L'] = op8;
vector<int> op9; op9.push_back(); op9.push_back(); op9.push_back(); Move['R'] = op9;
vector<int> op10; op10.push_back(); op10.push_back(); op10.push_back(); Move['U'] = op10;
vector<int> op11; op11.push_back(); op11.push_back(); op11.push_back(); Move['D'] = op11;
}
bool check(int x, int y)//判断越界
{
return x >= && x<n&&y >= && y<m;
}
int Get_F(Node a, int ex, int ey) //A*函数
{
return abs(a.x - ex) + abs(a.y - ey) + a.step;
}
char rotates(char str, int cnt) //得到rotates后按钮
{
char ST;
if (str == '-'){ cnt % ? ST = '|' : ST = '-'; }
else if (str == '|'){ cnt % ? ST = '-' : ST = '|'; }
else if (str == '^')
{
if (cnt % == ){ ST = '>'; }
else if (cnt % == ){ ST = 'v'; }
else if (cnt % == ){ ST = '<'; }
else{ ST = '^'; }
}
else if (str == '>')
{
if (cnt % == ){ ST = 'v'; }
else if (cnt % == ){ ST = '<'; }
else if (cnt % == ){ ST = '^'; }
else{ ST = '>'; }
}
else if (str == 'v')
{
if (cnt % == ){ ST = '<'; }
else if (cnt % == ){ ST = '^'; }
else if (cnt % == ){ ST = '>'; }
else{ ST = 'v'; }
}
else if (str == '<')
{
if (cnt % == ){ ST = '^'; }
else if (cnt % == ){ ST = '>'; }
else if (cnt % == ){ ST = 'v'; }
else { ST = '<'; }
}
else if (str == 'L')
{
if (cnt % == ){ ST = 'U'; }
else if (cnt % == ){ ST = 'R'; }
else if (cnt % == ){ ST = 'D'; }
else{ ST = 'L'; }
}
else if (str == 'R')
{
if (cnt % == ){ ST = 'D'; }
else if (cnt % == ){ ST = 'L'; }
else if (cnt % == ){ ST = 'U'; }
else{ ST = 'R'; }
}
else if (str == 'U')
{
if (cnt % == ){ ST = 'R'; }
else if (cnt % == ){ ST = 'D'; }
else if (cnt % == ){ ST = 'L'; }
else{ ST = 'U'; }
}
else if (str == 'D')
{
if (cnt % == ){ ST = 'L'; }
else if (cnt % == ){ ST = 'U'; }
else if (cnt % == ){ ST = 'R'; }
else{ ST = 'D'; }
}
else if (str == '+'){ ST = '+'; }
return ST;
}
bool checkMove(Node b, Node a,int op)//check B can to A?
{
char moveop = rotates(b.str, b.cnt);
if (op == )
{
for (int i = ; i < Move[moveop].size(); i++)
{
if (Move[moveop][i] == )
{
return true;
}
}
}
else if (op == )
{
for (int i = ; i < Move[moveop].size(); i++)
{
if (Move[moveop][i] == )
{
return true;
}
}
}
else if (op == )
{
for (int i = ; i < Move[moveop].size(); i++)
{
if (Move[moveop][i] == )
{
return true;
}
}
}
else
{
for (int i = ; i < Move[moveop].size(); i++)
{
if (Move[moveop][i] == )
{
return true;
}
}
}
return false;
}
int dfs(Node s, Node e)
{
//priority_queue<Node, vector<Node>, cmp >Q;
queue<Node>Q;
memset(vis, , sizeof(vis));
Q.push(s);
vis[s.x][s.y][s.cnt % ] = ;
while (!Q.empty())
{
Node Top = Q.front(); Q.pop();
//Node Top = Q.top(); Q.pop();
if (Top.x == e.x&&Top.y == e.y)
{
return Top.step;
}
Node next;
//rotates:
next.x = Top.x; next.y = Top.y; next.str = g[next.x][next.y]; next.cnt = Top.cnt + ; next.step = Top.step + ; next.f = Get_F(next, e.x, e.y);
if (vis[next.x][next.y][next.cnt % ] == )
{
vis[next.x][next.y][next.cnt % ] = ;
Q.push(next);
} //Move:
next.cnt = Top.cnt; next.str = g[next.x][next.y];
char op = rotates(next.str, next.cnt);
for (int i = ; i<Move[op].size(); i++)
{
next.x = Top.x + dist[Move[op][i]][];
next.y = Top.y + dist[Move[op][i]][];
if (check(next.x, next.y) && g[next.x][next.y] != '*'&&!vis[next.x][next.y][next.cnt % ])
{
next.str = g[next.x][next.y];
if (checkMove(next, Top,Move[op][i]))//检测是否B也能到A
{
vis[next.x][next.y][next.cnt % ] = ;
next.step = Top.step + ;
next.f = Get_F(next, e.x, e.y);
Q.push(next);
}
}
}
}
return -;
}
int main()
{
//#ifdef CYM
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
//#endif
init();
while (~scanf("%d%d", &n, &m)){
for (int i = ; i<n; i++)
{
scanf("%s", g[i]);
}
Node start, end;
scanf("%d%d", &start.x, &start.y);
start.x--; start.y--;
scanf("%d%d", &end.x, &end.y);
end.x--; end.y--;
start.step = , start.str = g[start.x][start.y], start.cnt = ; start.f = Get_F(start, end.x, end.y);
printf("%d\n", dfs(start, end));
}
return ;
}