CodeForces 540C Ice Cave （BFS）

http://codeforces.com/problemset/problem/540/C

 

 

 

Ice Cave

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

Submit Status Practice CodeForces 540C

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

题目大意：

m*n的冰宫，'X'代表破损，'.'代完好，你初始的位置为（r1, c1）,因为你是从上一层掉下来的，所以初始位置为'X'是破损的，你要从（r2, c2）的位置穿到下一层

你走过的地方冰块都会破损，所以处理（r2, c2）外是'X'的你都不能走，但（r2，c2）这个点你需要走两次，第一次为了破损他让他变为'X',第二次到他是通过他到下一层

BFS搜索，DFS会TLE

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<algorithm>

using namespace std;
const int N = ;
typedef __int64 ll;

char maps[N][N];
int d[][] = {{, }, {-, }, {, }, {, -}};
int m, n, f, r2, c2;

struct node
{
    int x, y;
};

bool BFS(int x, int y)
{
    queue<node>Q;
    node now, next;
    now.x = x;
    now.y = y;
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        for(int i =  ; i <  ; i++)
        {
            int a = next.x = now.x + d[i][];
            int b = next.y = now.y + d[i][];
            if(a == r2 && b == c2 && maps[a][b] == 'X')
                return true;
            if(a >=  && a < m && b >=  && b < n && maps[a][b] == '.')
            {
                if(maps[a][b] == '.')
                    maps[a][b] = 'X';//走过的地方，冰块会破损，将其变为'X'
                Q.push(next);
            }
        }
    }
    return false;
}

int main()
{
    int r1, c1;
    while(~scanf("%d%d", &m, &n))
    {
        f = ;
        for(int i =  ; i < m ; i++)
            scanf("%s", maps[i]);
        scanf("%d%d", &r1, &c1);
        scanf("%d%d", &r2, &c2);
        r1--;
        c1--;
        r2--;
        c2--;
        if(BFS(r1, c1))
            printf("YES\n");
        else
            printf("NO\n");
    }
    return ;
}