| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8046 | Accepted: 3518 |
Description
One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)
As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.
Input
Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').
You may assume that the maze exit is always reachable from the start point.
Output
Sample Input
2
8 8
########
#......#
#.####.#
#.####.#
#.####.#
#.####.#
#...#..#
#S#E####
9 5
#########
#.#.#.#.#
S.......E
#.#.#.#.#
#########
Sample Output
37 5 5
17 17 9
#include<iostream>
#include<stdio.h>
using namespace std;
int w,h,ex,ey,sx,sy;
int map[100][100],can[100][100];
struct vid{ int x,y,step;
}queue[5000];
int zan[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
int dirl[4][2]={0,-1,-1,0,0,1,1,0},dirr[4][2]={0,1,1,0,0,-1,-1,0};
int dfsr(int dstep,int x,int y,int di )
{
int i,temp1,temp2; for(i=0;i<4;i++)
{
temp1=x+dirr[i][0];
temp2=y+dirr[i][1]; if((temp1>=0)&&(temp1<h)&&(temp2>=0)&&(temp2<w))
{
if(dfsr(dstep+1,x+dirr[i][0],y+dirr[i][1],i))
return dstep;
else
return 0;
} }
return 0; }
int dfsl(int dstep,int x,int y,int di )
{ int i,temp1,temp2; for(i=0;i<4;i++)
{
temp1=x+dirl[i][0];
temp2=y+dirl[i][1]; if((temp1>=0)&&(temp1<h)&&(temp2>=0)&&(temp2<w))
{
if(dfsl(dstep+1,temp1,temp2,i))
return dstep;
else
return 0;
} } return 0;
}
int bfs()
{ if (sx == ex && sy == ey)
{ return 1;
}
int t,ww,x,y,t1,t2;
t=ww=1;
queue[t].x=sx;
queue[t].y=sy;
queue[t].step=0;
can[sx][sy]=1;
while(t<=ww&&!can[ex][ey])
{ x=queue[t].x;
y=queue[t].y;
for(int i=0;i<4;i++)
if((!map[x+zan[i][0]][y+zan[i][1]])&&(!can[x+zan[i][0]][y+zan[i][1]]))
{ t1=x+zan[i][0];
t2=y+zan[i][1];
if(t1>=0&&(t1<h)&&(t2>=0)&&(t2<w)&&(!map[t1][t2])&&(!can[t1][t2]))
{ queue[++ww].x=t1;
queue[ww].y=t2;
queue[ww].step=queue[t].step+1;
can[t1][t2]=1;
}
}
t++;
}
return queue[ww].step+1;
}
int main ()
{
int t;
char c;
scanf("%d",&t);
getchar();
while(t--)
{ scanf("%d%d",&w,&h);
getchar();
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
can[i][j]=0;
c=getchar();
if(c=='#')
map[i][j]=1;
else if(c=='.')
map[i][j]=0;
else if(c=='S')
{
map[i][j]=0;
sx=i,sy=j;
}
else
if(c=='E')
{
map[i][j]=0;
ex=i;ey=j;
}
}
getchar();
}
// init();
// dfsr();
// printf("%d ",bfs());
printf("%d %d %d\n",dfsl(0,sx,sy,0)+1,dfsr(0,sx,sy,0)+1,bfs());
} return 0;
}