C. Ice Cave
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 test(s)
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
Note
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.
链接：http://codeforces.com/problemset/problem/540/C
题意：X是坑，‘.‘是冰层，冰层走过一次后变成坑。问能不能从 sx，sy开始走， 在ex，ey处掉入坑中。
做法：bfs一下，X处如果不是终点，那肯定是不能走的。 如果是终点，且是X，就直接返回可以。 如果是‘.’那都可以走。走完之后把这个格子变成‘X’。复杂度为地图大小*2。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
int dir[4][2]={1,0,-1,0,0,1,0,-1};
char mp[510][510];
int n,m;
int ex,ey;
int flag;
struct point
{
int x,y;
};
int ok(int x,int y)
{
if(x<1||x>n||y<1||y>m)
return 0;
return 1;
}
int bfs(int x,int y)
{
queue<point> q;
point sta,nw,nxt,ep;
ep.x=ex;
ep.y=ey;
sta.x=x,sta.y=y;
q.push(sta);
while(!q.empty())
{
nw=q.front();
q.pop();
//printf("%d %d\n",nw.x,nw.y);
for(int i=0;i<4;i++)
{
nxt.x=nw.x+dir[i][0];
nxt.y=nw.y+dir[i][1];
if(ok(nxt.x,nxt.y))
{
if(nxt.x==ep.x&&nxt.y==ep.y)
{
if(mp[nxt.x][nxt.y]=='.')
mp[nxt.x][nxt.y]='X';
else
return 1;
q.push(nxt);
}
else if(mp[nxt.x][nxt.y]=='.')
{
mp[nxt.x][nxt.y]='X';
q.push(nxt);
}
}
}
}
return 0;
}
int main()
{
int sx,sy;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%s",mp[i]+1);
scanf("%d%d",&sx,&sy);
scanf("%d%d",&ex,&ey);
flag=0;
if(bfs(sx,sy))
puts("YES");
else
puts("NO");
}
return 0;
}