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.
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 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.
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.
题目大意：
主人公有一个起点，有一个终点，各个形状表示各种不同的门，从一个点能够走到另一个点的要求是两个点之间都有门。
每一单位时间可以选择将所有门旋转90度，或者是移动到相邻的一个格子上。
思路：
显然每个点有四种状态，那么设定vis【i】【j】【k】表示到达点（i，j）处，状态为k的最小步数。
那么过程模拟，维护最小值。
每种状态都要枚举到。
比较复杂，谨慎点即可。
Ac代码：
#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
using namespace std;
struct node
{
int x,y;
int contz;
int step;
friend bool operator <(node a,node b)
{
return a.step>b.step;
}
} now,nex;
int fx[4]= {-1,1,0,0};
int fy[4]= {0,0,-1,1};
char a[1515][1515];
int vis[1515][1515][5];
int n,m,u,l,d,r;
int u2,l2,d2,r2;
int judge(int x,int y)
{
if(x>=0&&x<n&&y>=0&&y<m&&a[x][y]!='*')return 1;
else return 0;
}
int aa[10];
int bb[10];
void initaa()
{
if(a[now.x][now.y]=='+')aa[0]=1,aa[1]=1,aa[2]=1,aa[3]=1;
if(a[now.x][now.y]=='-')aa[0]=0,aa[1]=1,aa[2]=0,aa[3]=1;
if(a[now.x][now.y]=='|')aa[0]=1,aa[1]=0,aa[2]=1,aa[3]=0;
if(a[now.x][now.y]=='^')aa[0]=1,aa[1]=0,aa[2]=0,aa[3]=0;
if(a[now.x][now.y]=='>')aa[0]=0,aa[1]=1,aa[2]=0,aa[3]=0;
if(a[now.x][now.y]=='v')aa[0]=0,aa[1]=0,aa[2]=1,aa[3]=0;
if(a[now.x][now.y]=='<')aa[0]=0,aa[1]=0,aa[2]=0,aa[3]=1;
if(a[now.x][now.y]=='L')aa[0]=1,aa[1]=1,aa[2]=1,aa[3]=0;
if(a[now.x][now.y]=='R')aa[0]=1,aa[1]=0,aa[2]=1,aa[3]=1;
if(a[now.x][now.y]=='U')aa[0]=0,aa[1]=1,aa[2]=1,aa[3]=1;
if(a[now.x][now.y]=='D')aa[0]=1,aa[1]=1,aa[2]=0,aa[3]=1;
}
void initbb()
{
if(a[nex.x][nex.y]=='+')bb[0]=1,bb[1]=1,bb[2]=1,bb[3]=1;
if(a[nex.x][nex.y]=='-')bb[0]=0,bb[1]=1,bb[2]=0,bb[3]=1;
if(a[nex.x][nex.y]=='|')bb[0]=1,bb[1]=0,bb[2]=1,bb[3]=0;
if(a[nex.x][nex.y]=='^')bb[0]=1,bb[1]=0,bb[2]=0,bb[3]=0;
if(a[nex.x][nex.y]=='>')bb[0]=0,bb[1]=1,bb[2]=0,bb[3]=0;
if(a[nex.x][nex.y]=='v')bb[0]=0,bb[1]=0,bb[2]=1,bb[3]=0;
if(a[nex.x][nex.y]=='<')bb[0]=0,bb[1]=0,bb[2]=0,bb[3]=1;
if(a[nex.x][nex.y]=='L')bb[0]=1,bb[1]=1,bb[2]=1,bb[3]=0;
if(a[nex.x][nex.y]=='R')bb[0]=1,bb[1]=0,bb[2]=1,bb[3]=1;
if(a[nex.x][nex.y]=='U')bb[0]=0,bb[1]=1,bb[2]=1,bb[3]=1;
if(a[nex.x][nex.y]=='D')bb[0]=1,bb[1]=1,bb[2]=0,bb[3]=1;
}
void Bfs(int sx,int sy,int ex,int ey)
{
for(int i=0;i<=1444;i++)
{
for(int j=0;j<=1444;j++)
{
for(int k=0;k<=4;k++)
{
vis[i][j][k]=0x3f3f3f3f;
}
}
}
sx--,sy--,ex--,ey--;
priority_queue<node >s;
now.x=sx;
now.y=sy;
now.step=0;
now.contz=0;
s.push(now);
vis[now.x][now.y][0]=0;
while(!s.empty())
{
now=s.top();
if(now.x==ex&&now.y==ey)
{
printf("%d\n",now.step);
return ;
}
s.pop();
for(int i=0;i<4;i++)
{
nex.x=now.x+fx[i];
nex.y=now.y+fy[i];
if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m&&a[nex.x][nex.y]!='*')
{
initaa();initbb();
int pre=now.contz%4;
for(int i=0;i<pre;i++)
{
int tmp=aa[3];
aa[3]=aa[2];aa[2]=aa[1];aa[1]=aa[0];aa[0]=tmp;
tmp=bb[3];
bb[3]=bb[2];bb[2]=bb[1];bb[1]=bb[0];bb[0]=tmp;
}
for(int j=0;j<4;j++)
{
if(i==0)//u
{
if(aa[0]==1&&bb[2]==1)
{
nex.step=j+1+now.step;
nex.contz=now.contz+j;
if(vis[nex.x][nex.y][nex.contz%4]>nex.step)
{
vis[nex.x][nex.y][nex.contz%4]=nex.step;
s.push(nex);
}
}
}
if(i==1)//d
{
if(aa[2]==1&&bb[0]==1)
{
nex.step=j+1+now.step;
nex.contz=now.contz+j;
if(vis[nex.x][nex.y][nex.contz%4]>nex.step)
{
vis[nex.x][nex.y][nex.contz%4]=nex.step;
s.push(nex);
}
}
}
if(i==2)//l
{
if(aa[3]==1&&bb[1]==1)
{
nex.step=j+1+now.step;
nex.contz=now.contz+j;
if(vis[nex.x][nex.y][nex.contz%4]>nex.step)
{
vis[nex.x][nex.y][nex.contz%4]=nex.step;
s.push(nex);
}
}
}
if(i==3)//r
{
if(aa[1]==1&&bb[3]==1)
{
nex.step=j+1+now.step;
nex.contz=now.contz+j;
if(vis[nex.x][nex.y][nex.contz%4]>nex.step)
{
vis[nex.x][nex.y][nex.contz%4]=nex.step;
s.push(nex);
}
}
}
int tmp=aa[3];
aa[3]=aa[2];aa[2]=aa[1];aa[1]=aa[0];aa[0]=tmp;
tmp=bb[3];
bb[3]=bb[2];bb[2]=bb[1];bb[1]=bb[0];bb[0]=tmp;
}
}
}
}
/*
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
for(int k=0;k<=4;k++)
{
printf("%d ",vis[i][j][k]);
}
}
}
printf("\n");
*/
printf("-1\n");
return ;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
int x1,y1,x2,y2;
memset(vis,0,sizeof(vis));
for(int i=0; i<n; i++)scanf("%s",a[i]);
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
Bfs(x1,y1,x2,y2);
}
}
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