/*816 - Abbott‘s Revenge---代码完全参考刘汝佳算法入门经典---strchr() 用来查找某字符在字符串中首次出现的位置，其原型为：char * strchr (const char *str, int c)---BFS求最短路--*/#define _CRT_SECURE_NO_DEPRECATE#include<iostream>#include<string.h>#include<queue>#include<algorithm>using namespace std;const int maxn = 10;struct Node{	int r, c, dir;	Node(int a=0, int b=0, int c=0) :r(a), c(b), dir(c){}};int dis[maxn][maxn][4];//dis[r][c][dir]保存状态(r,c,dir)到初始状态的距离int id[256];Node path[maxn][maxn][4]; //path[r][c][dir]保存了状态(r,c,dir)在BFS树中的父节点int has[maxn][maxn][4][3];//has[r][c][dir][turn]，表示当前处在(r,c,dir)状态是否可以向turn转弯int r0, c0, r1, c1, r2, c2, dir;const int dr[] = { -1, 0, 1, 0 };const int dc[] = { 0, 1, 0, -1 };//计算出下一个节点Node walk(Node&u, int turn){	//首先计算出下一步的朝向	int dir = u.dir;	if (turn == 1)dir = (dir + 3) % 4; //左转，逆时针	if (turn == 2)dir = (dir + 1) % 4;//右转，顺时针	return Node(u.r + dr[dir], u.c + dc[dir], dir);}bool insid(int r, int c){	return r >= 1 && c >= 1 && r <= 9 && c <= 9;}void print_ans(Node u){	vector<Node>vec;	while (dis[u.r][u.c][u.dir] != 0){		vec.push_back(u);		u = path[u.r][u.c][u.dir];	}	vec.push_back(u);	vec.push_back(Node(r0, c0, dir));	int cnt = 0;	printf("  ");	for (int i = vec.size() - 1; i >= 0; i--,cnt++){		if (cnt){			if (cnt% 10 == 0)printf("\n  ");			else printf(" ");		}		printf("(%d,%d)", vec[i].r, vec[i].c);	}	printf("\n");}void bfs(){	queue<Node>Q;	Node u(r1, c1, dir);	Q.push(u);	memset(dis, -1, sizeof(dis));	dis[r1][c1][dir] = 0;	while (!Q.empty()){		u = Q.front(); Q.pop();		if (u.r == r2&&u.c == c2){ print_ans(u); return; }		for (int i = 0; i < 3; i++){			Node v = walk(u, i);			if (has[u.r][u.c][u.dir][i] && insid(v.r, v.c) && dis[v.r][v.c][v.dir] < 0){				path[v.r][v.c][v.dir] = u;				dis[v.r][v.c][v.dir] = dis[u.r][u.c][u.dir] + 1;				Q.push(v);			}		}	}	printf("  No Solution Possible\n");}int main(){	//01234代表NESW，顺时针方向	id[‘N‘] = 0;	id[‘E‘] = 1;	id[‘S‘] = 2;	id[‘W‘] = 3;	//012代表转向	id[‘F‘] = 0;	id[‘L‘] = 1;	id[‘R‘] = 2;	char s1[21], s2[21];	while (scanf("%s", s1)&&strcmp(s1,"END")){		printf("%s\n", s1);		scanf("%d%d%s%d%d", &r0, &c0, s2, &r2, &c2);		dir = id[s2[0]];		r1 = r0 + dr[dir];		c1 = c0 + dc[dir];		memset(has, 0, sizeof(has));		int r, c;		while (scanf("%d", &r) && r){			scanf("%d", &c);			while (scanf("%s", s1) && strcmp(s1, "*")){				for (int i = 1; i < strlen(s1); i++){					has[r][c][id[s1[0]]][id[s1[i]]] = 1;				}			}		}		bfs();	}	return 0;}