本題和一般的迷宮問題本質上是一樣的，但是由於朝向問題，所以需要一個三元組儲存node(r, c, dir)。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 10;
const char* dirs = "NESW";
const char* turns = "FLR";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1}; 
int d[maxn][maxn][4];
int r0, c0, dir, r1, c1, r2, c2;
int hasEdge[maxn][maxn][4][3];
struct node{
	int r, c;
	int dir;
	node(int r = 0, int c = 0, int dir = 0):r(r), c(c), dir(dir){}
};

node p[maxn][maxn][4];
int dirId(char ch){
	return strchr(dirs, ch) - dirs;
}
int turnId(char ch){
	return strchr(turns, ch) - turns;
}
node walk(const node &u, int turn){
	int dir = u.dir;
	if(turn == 1) dir = (dir + 3) % 4;
	if(turn == 2) dir = (dir + 1) % 4;
	return node(u.r + dr[dir], u.c + dc[dir], dir);
}
void print(node u){
	vector<node> nodes;
	while(d[u.r][u.c][u.dir] != 0){
		nodes.push_back(u);
		u = p[u.r][u.c][u.dir];
	}
	nodes.push_back(u);
	nodes.push_back(node(r0, c0, dir));
	int cnt = 0;
	for(int i = nodes.size()-1; i >= 0; i--) {
	    if(cnt % 10 == 0) printf(" ");
	    printf(" (%d,%d)", nodes[i].r, nodes[i].c);
	    if(++cnt % 10 == 0) printf("\n");
	}
    if(nodes.size() % 10 != 0) printf("\n");
}
bool read(){
	char s1[99], s2[99];
	if(scanf("%s%d%d%s%d%d", s1, &r0, &c0, s2, &r2, &c2) != 6) return false;
	dir = dirId(s2[0]);
	printf("%s\n", s1);
	r1 = r0 + dr[dir];
	c1 = c0 + dc[dir];
	int r, c;
	memset(hasEdge, 0, sizeof(hasEdge));
	while(scanf("%d", &r) == 1 && r){
		scanf("%d", &c);
		while(scanf("%s", s1) == 1 && s1[0] != '*'){
			for(int i = 1; i < strlen(s1); ++i){
				hasEdge[r][c][dirId(s1[0])][turnId(s1[i])] = 1;
			}
		}
	}
	return true;
}
bool inside(int r, int c){
	return r > 0 && r <= 9 && c > 0 && c <= 9;
}

void solve(){
	node u(r1, c1, dir);
	memset(d, -1, sizeof(d));
	queue<node> q;
	q.push(u);
	d[r1][c1][dir] = 0;
	while(!q.empty()){
		u = q.front(); q.pop();
		if(u.r == r2 && u.c == c2) {print(u); return;}
		for(int i = 0; i < 3; ++i){
			node v = walk(u, i);
			if(hasEdge[u.r][u.c][u.dir][i] && inside(v.r, v.c) && d[v.r][v.c][v.dir] < 0){
				d[v.r][v.c][v.dir] = d[u.r][u.c][u.dir] + 1;
				p[v.r][v.c][v.dir] = u;
				q.push(v);
			}
		}
	}
	printf("  No Solution Possible\n");
}
int main(){
	while(read()){
		solve();
	}
	return 0;
}