Ice Cave--Codeforces(广度优先搜索)

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
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.

首先跟大家解释一下题目大意：假如你在玩一个游戏，游戏的规则是这样的，你从上层冰冻掉下来掉到一个位置上，此处的冰是裂的（比较特殊，虽然是裂开的，但你不会掉落下去），冰块分为两种，一种是’. ‘,代表此时的冰是完整的，另一种是“X”，代表此时的冰是裂开的，假如你踩在一个裂开的冰块上面，然后你会从这块冰块上掉下去，现在给你开始的位置和进入下层的位置。问你是否能到达下层，如果能的话输出“YES”，否则的话输出“NO”；本题的较大的坑点是假如终点处是“.”的话你需要判断周围是否还有另外一个还没有走过的点，如果有的话先到这个点上去，然后再回到终点。
我用广搜处理了这个问题，一开始不会弄，后来WRONG 了很多发以后找到规律了。然后就A掉了。

下面附上AC代码；

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

int dir[4][2]= {{1,0},{-1,0},{0,-1},{0,1}};
int n,m;
char a[505][505];
struct node
{
    int x,y;
} fr,ne;
int book[505][505];
int sx,sy,ex,ey;
int bfs(int sx,int sy,int ex,int ey)
{
    memset(book,0,sizeof(book));
    book[sx][sy]=1;
    queue<node>q;
    ne.x=sx,ne.y=sy;
    q.push(ne);
    while(!q.empty())
    {
        fr=q.front();
        q.pop();
        for(int i=0; i<4; i++)
        {
            int fx=fr.x+dir[i][0];
            int fy=fr.y+dir[i][1];
            if(fx==ex&&fy==ey&&a[fx][fy]=='X')
            {
                return 1;
            }
            if(fx>=0&&fx<n&&fy>=0&&fy<m&&a[fx][fy]=='.'&&book[fx][fy]==0)
            {
                book[fx][fy]=1;
                a[fx][fy]='X';
                ne.x=fx;
                ne.y=fy;
                q.push(ne);
            }
        }
    }
    return 0;
}

int main()
{
    while(~scanf("%d%d\n",&n,&m))
    {
        for(int i=0; i<n; i++)
        {
            scanf("%s",&a[i]);
        }
        scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
        sx--,sy--,ex--,ey--;
        int re=bfs(sx,sy,ex,ey);
        if(re==1)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}