• 题意
    • 输入蜂巢图,求最短路径
  • 题解
    • 一开始一直在想怎么输入,然后建图解决问题
    • 但实际上并不需要重新建图,我么发现,每个格子的中间位置,都满足按照六个方向通行的规律,两个格子之间能够通行,只需要满足1个条件
      • 两个格子中间是空格
    • 所以我们只需要把六个方向分别判定一次,bfs即可
    • 图很大,不能memset(vis,0,sizeof(vis))
  • 代码
    • #include<algorithm>
      #include <iostream>
      #include  <fstream>
      #include  <cstdlib>
      #include  <cstring>
      #include  <cassert>
      #include   <cstdio>
      #include   <vector>
      #include   <string>
      #include    <cmath>
      #include    <queue>
      #include    <stack>
      #include      <set>
      #include      <map>
      using namespace std;
      #define P(a,b,c) make_pair(a,make_pair(b,c))
      #define rep(i,a,n) for (int i=a;i<=n;i++)
      #define per(i,a,n) for (int i=n;i>=a;i--)
      #define CLR(vis) memset(vis,0,sizeof(vis))
      #define MST(vis,pos) memset(vis,pos,sizeof(vis))
      #define pb push_back
      //#define mp make_pair
      #define all(x) (x).begin(),(x).end()
      #define fi first
      #define se second
      #define SZ(x) ((int)(x).size())
      typedef pair<int,pair<int,int> >pii;
      typedef long long ll;
      typedef unsigned long long ull;
      const ll mod = 1000000007;
      const int INF = 0x3f3f3f3f;
      ll gcd(ll a, ll b) {
      	return b ? gcd(b, a%b) : a;
      }
      template<class T>inline void gmax(T &A, T B) {
      	(A<B)&&(A=B);//if(B>A)A=B;
      }
      template<class T>inline void gmin(T &A, T B) {
      	(A>B)&&(A=B);//if(B<A)A=B;
      }
      template <class T>
      inline bool scan(T &ret) {
      	char c;
      	int sgn;
      	if(c=getchar(),c==EOF) return 0; //EOF
      	while(c!='-'&&(c<'0'||c>'9')) c=getchar();
      	sgn=(c=='-')?-1:1;
      	ret=(c=='-')?0:(c-'0');
      	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
      	ret*=sgn;
      	return 1;
      }
      inline void outt(ll x) {
      	if(x>9) outt(x/10);
      	putchar(x%10+'0');
      }
      
      const int maxn = 10005;
      
      char g[maxn][maxn];
      int vis[maxn][maxn];
      struct ss{
      	int x,y,step;
      	ss(int _x=0,int _y=0,int _step=0):x(_x),y(_y),step(_step){
      	}
      };
      
      int dir[6][2] = {1,3,1,-3,-1,3,-1,-3,2,0,-2,0};
      int stx,sty;
      int edx,edy;
      int n,m;
      bool ok(int x,int y){
      	if(x>0&&x<=n&&y>0&&y<=m&&vis[x][y]==0&&(g[x][y]==' '||g[x][y]=='T'))return true;
      	return false;
      }
      
      int bfs(){
      	queue<ss>q;
      	while(!q.empty()) q.pop();
      	q.push(ss{stx,sty,1});
      	vis[stx][sty] = 1;
      	while(!q.empty()){
      		int x=q.front().x, y=q.front().y, step=q.front().step;
      		if(x==edx&&y==edy) return step;
      		q.pop();
      		rep(i,0,5){
      			int xx = x+dir[i][0], yy = y+dir[i][1];
      			if(ok(xx,yy)){
      				vis[xx][yy]=1;
      				int xxx=xx+dir[i][0], yyy = yy+dir[i][1];
      				if(ok(xxx,yyy)){
      					vis[xxx][yyy]=1;
      					q.push(ss{xxx,yyy,step+1});
      				}
      			}
      		}
      	}
      	return -1;
      }
      
      
      int main(){
      	int t;
      	scanf("%d", &t);
      	while(t--){
      		scanf("%d%d", &n,&m);
      		getchar();
      		n = n*4+3, m = m*6+3;
      		rep(i,1,n){
      			gets(g[i]);
      			rep(j,1,m){
      				if(g[i][j]=='\0'){
      					break;
      				}
      				vis[i][j]=0;
      				if(g[i][j] == 'S')
      					stx = i, sty = j;
      				if(g[i][j] == 'T')
      					edx = i, edy = j;
      			}
      		}
      		int ans = bfs();
      		printf("%d\n", ans);
      	}
      	return 0;
      }

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