SunnyPig is a pig who is much cleverer than any other pigs in the pigpen. One sunny morning, SunnyPig wants to go out of the pigpen to date Mrs. Snail, his beloved. However, it’s terribly tough for a pig to go out of the pigpen because the pigpen is divided into m * n grids with fences which pigs cannot go across. Luckily, there are some doors unlocked on the fences so that SunnyPig can push them open with his nose. Since SunnyPig is a pig, no matter how clever he is, he can never walk upright like human beings. As a result, SunnyPig is not able to pull any doors. Now give you the map of the pigpen, please calculate the fewest number of doors SunnyPig should push to go out of the pigpen.
The first line there is a number T (0 < T < 100), denoting the number of the test case. The first line of each test case has only two numbers: m, n. The following 2*m+1 lines describe the pigpen. Each line has 2*n+1 characters. ’*’ represents a cross point of two fences. ’O’ represents the initial position SunnyPig. ’-’ and ‘|’ represent fences without doors. ’N’, ’S’, ’W’, ’E’ represent the doors SunnyPig can push in the direction of north, south, west and east respectively. And the character of a space represents the place where SunnyPig can go through.
Output the fewest number of doors SunnyPig should push to go out of the pigpen, in other words, the fewest number of doors SunnyPig should push to go out of the border of these grids. If SunnyPig cannot go out of the pigpen, output -1. Each case, a single line.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
int m,n;
char mp[2005][2005];
int dis[4][2]= {{1,0},{-1,0},{0,-1},{0,1}};
char ch[4]={‘S‘,‘N‘,‘W‘,‘E‘};
struct node
{
int x;
int y;
int tim;
};
queue<node> q;
node now,ww;
int sx,sy;
int bfs(int xx,int yy)
{
while(!q.empty())
q.pop();
now.x=xx;
now.y=yy;
now.tim=0;
mp[xx][yy]=‘2‘;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
//cout<<now.x<<now.y<<endl;
if(now.x==0||now.x==2*(m+1)||now.y==0||now.y==2*(n+1))
return now.tim;
for(int i=0; i<=3; i++)
{
//cout<<"****"<<endl;
int s1=now.x+dis[i][0];
int s2=now.y+dis[i][1];
if(s1>=0&&s1<=2*(m+1)&&s2>=0&&s2<=2*(n+1)&&(mp[s1][s2]==‘1‘||mp[s1][s2]==‘ ‘||mp[s1][s2]==ch[i]))
{
if(mp[s1][s2]==ch[i])
ww.tim=now.tim+1;
else
ww.tim=now.tim;
ww.x=s1;
ww.y=s2;
mp[s1][s2]=‘2‘;
q.push(ww);
}
}
}
return -1;
}
int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
for(int i=1; i<=t; i++)
{
memset(mp,0,sizeof(mp));
scanf("%d%d",&m,&n);
for(int j=0; j<=2*m+2; j++)
{
for(int k=0; k<=2*n+2; k++)
{
if(j==0||j==2*(m+1)||k==0||k==2*(n+1))
mp[j][k]=‘1‘;
else
{
scanf("%c",&mp[j][k]);
if(mp[j][k]==‘O‘)
{
sx=j;
sy=k;
}
}
}
char ke[2];
if(j!=2*m+2)
gets(ke);//用getchar 就wa 有坑点
}
cout<<bfs(sx,sy)<<endl;
}
}
return 0;
}
