CF 354 D 迷宫与门的旋转 BFS +状态压缩 一定要回头看看

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 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

Note

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

题意：给你一个n*m的地图，然后地图有很多标志如题所述

然后他每秒钟要么可以穿过门，要么可以使得所有门都顺时针转动90°

如果你要从A到B，那么从B也必须能够到达A

问你从起点到终点的最短时间是多少

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-;
const int  inf =0x7f7f7f7f;
const double pi=acos(-);
const int maxn=+;

int ans=inf;
int dx[]={-,,,};
int dy[]={,,,-};
int n,m,sx,sy,tx,ty;
char s[maxn][maxn];
int a[maxn][maxn],vis[maxn][maxn][];

void init(int x,int y)
{
    char c=s[x][y];
    if(c=='+')
        a[x][y]=;
    else if(c=='-')
        a[x][y]=;
    else if(c=='|')
        a[x][y]=;
    else if(c=='^')
        a[x][y]=;
    else if(c=='>')
        a[x][y]=;
    else if(c=='v')
        a[x][y]=;
    else if(c=='<')
        a[x][y]=;
    else if(c=='L')
        a[x][y]=;
    else if(c=='U')
        a[x][y]=;
    else if(c=='R')
        a[x][y]=;
    else if(c=='D')
        a[x][y]=;
    else if(c=='*')
        a[x][y]=;
}

struct node{
    int x,y,dir;
    ll dis;
    node(int a,int b,int c,ll d):x(a),y(b),dir(c),dis(d){};
};

bool legal(int x,int y)
{
   return x>=&&x<=n&&y>=&&y<=m;
}

ll bfs()
{
    node fir(sx,sy,,);
    queue<node> q;
    q.push(fir);
    vis[sx][sy][]=;
    while(q.size())
    {
        node u=q.front();q.pop();
        int ux=u.x,uy=u.y;
        if(ux==tx&&uy==ty) return u.dis;
        for(int i=;i<;i++)
        {
            int vx=u.x+dx[i];
            int vy=u.y+dy[i];
            if(!legal(vx,vy)) continue;
            if(vis[vx][vy][u.dir]) continue;
            int k1=(a[ux][uy]>>((i+-u.dir)%))&;
            int k2=(a[vx][vy]>>((i++-u.dir)%))&;
            if(k1&&k2)
                 {
                     q.push(node(vx,vy,u.dir,u.dis+));
                     vis[vx][vy][u.dir]=;
                 }
        }

        if(!vis[ux][uy][(u.dir+1)%4])
                 {
                    q.push(node(ux,uy,(u.dir+1)%4,u.dis+1));
                    vis[ux][uy][(u.dir+1)%4]=1;
                 }
    }
    return -;
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        MM(vis,);MM(a,);
        for(int i=;i<=n;i++)
        {
            scanf("%s",s[i]+);
            for(int j=;j<=m;j++)
                    init(i,j);
        }

        scanf("%d %d",&sx,&sy);
        scanf("%d %d",&tx,&ty);

        printf("%lld\n",bfs());
    }
    return ;
}

分析：好难的搜索啊,,,,,应该是做过的最难的一个了，，回头复习下

把红色的部分换成如下形式也错了，，已难蠢，，回头好好理清下思路

for(int i=;i<=;i++)
            if(!vis[ux][uy][(u.dir+i)%])
                 {
                    q.push(node(ux,uy,(u.dir+i)%,u.dis+i));
                    vis[ux][uy][(u.dir+i)%]=;
                 }

错误的原因在与BFS时，是当前状态转移到下一个状态而用这个代码的话，则也转移到了下下个状态

就错了