Codeforces 301_div.2_Ice Cave（BFS走冰块）

Ice Cave

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.

The level of the cave where you are is a rectangular square grid of n rows and m columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.

Let's number the rows with integers from 1 to n from top to bottom and the columns with integers from 1 to m from left to right. Let's denote a cell on the intersection of the r-th row and the c-th column as (r, c).

You are staying in the cell (r₁, c₁) and this cell is cracked because you've just fallen here from a higher level. You need to fall down through the cell (r₂, c₂) since the exit to the next level is there. Can you do this?

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 500) — the number of rows and columns in the cave description.

Each of the next n lines describes the initial state of the level of the cave, each line consists of m characters "." (that is, intact ice) and "X" (cracked ice).

The next line contains two integers, r₁ and c₁ (1 ≤ r₁ ≤ n, 1 ≤ c₁ ≤ m) — your initial coordinates. It is guaranteed that the description of the cave contains character 'X' in cell (r₁, c₁), that is, the ice on the starting cell is initially cracked.

The next line contains two integers r₂ and c₂ (1 ≤ r₂ ≤ n, 1 ≤ c₂ ≤ m) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one.

Output

If you can reach the destination, print 'YES', otherwise print 'NO'.

Sample Input

Input

4 6
X...XX
...XX.
.X..X.
......
1 6
2 2

Output

YES

Input

5 4
.X..
...X
X.X.
....
.XX.
5 3
1 1

Output

NO

Input

4 7
..X.XX.
.XX..X.
X...X..
X......
2 2
1 6

Output

YES

Hint（看下面）

In the first sample test one possible path is:

After the first visit of cell (2, 2) the ice on it cracks and when you step there for the second time, your character falls through the ice as intended.

BFS搜索！！很明显了！

要注意的是冰块踩过会变成X，下一次踩就会掉下去。

题解：pair是一种模板类型，其中包含两个数据值，两个数据的类型可以相同可以不同 ，由于pair类型的使用比较繁琐，因为如果要定义多个形同的pair类型的时候，可以时候typedef简化声明：
typedef pair<int, int>p;p中含有2个整型数

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
typedef pair<int, int> p;
int dx[]= { , -, ,  };
int dy[]= { , , , - };
int n,m;
int r1,c1,r2,c2;
char a[][];
int bfs()
{
    queue<p> q;
    q.push(p(r1,c1));
    a[r1][c1]='X';                //第一步上去冰块已碎
    while(!q.empty())
    {
        r1=q.front().first;
        c1=q.front().second;
        q.pop();
        for(int i=; i<; i++)
        {
            int xx=r1+dx[i];
            int yy=c1+dy[i];
            if( xx<||xx>=n||yy<||yy>=m)
                continue;
            if(a[xx][yy]=='X')
            {
                                     //到达点也会碎
                if(xx==r2 && yy==c2)
                    return ;
                continue;
            }
            a[xx][yy]='X';     //走了之后就变成易碎的冰块
            q.push(p(xx,yy));
        }
    }
    return ;
}
int main()
{
    int ans;
    while(cin>>n>>m)
    {
        for(int i=; i<n; i++)
            cin>>a[i];
        cin>>r1>>c1>>r2>>c2;
        r1--;       //用的二维数组是从0~n-1，而输入的是1~n,所以都要减1
        c1--;
        r2--;
        c2--;
        ans=bfs();
        if(ans)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return ;
}