Codeforces Round #301 (Div. 2) C. Ice Cave BFS

C. Ice Cave

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

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

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

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

 

Sample Output

YES

 

HINT

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.

题意

有一个矩形区域，一开始你在一个地方，你只能走.的位置，走了之后.就会变成X，然后你必须让终点变成x，然后再走上去

问你可不可行

题解：

BFS搞一搞就好了，裸题= =

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //§ß§é§à§é¨f§³
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

string s[1000];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
struct node
{
    int x,y;
    int t;
};
int vis[1000][1000];
int main()
{
    int n=read(),m=read();
    for(int i=0;i<n;i++)
        cin>>s[i];
    node st,ed;
    cin>>st.x>>st.y;
    st.x--,st.y--;
    st.t=0;
    cin>>ed.x>>ed.y;
    ed.x--,ed.y--;
    queue<node> q;
    q.push(st);
    while(!q.empty())
    {
        node now=q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            node next=now;
            next.x+=dx[i];
            next.y+=dy[i];
            next.t++;
            if(next.x<0||next.x>=n||next.y<0||next.y>=m)
                continue;
            if(next.x==ed.x&&next.y==ed.y&&s[next.x][next.y]=='X')
            {
                printf("YES\n");
                return 0;
            }
            if(s[next.x][next.y]=='X')
                continue;
            q.push(next);
            s[next.x][next.y]='X';
        }
    }
    printf("NO\n");
}