uva816

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=35521

/* 1：将方向用数组来表示，可方便编写代码。 2：注意题意，初始状态是刚刚离开入口 3：注意输出的格式，最后如果不满10个坐标就结束记得换行 */
#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <bits/stdc++.h>

using namespace std;

const char * dir_s = "NESW";
const char * turn_s = "FLR";

const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};

int dirToInt(char c) {
    return strchr(dir_s, c) - dir_s;
}

int turnToInt(char c) {
    return strchr(turn_s, c) - turn_s;
}

struct Node {
    int r, c, dir;
    Node(int r = 0, int c = 0, int dir = 0) : r(r), c(c), dir(dir) {}
};

int r0, c0, dir, r1, c1, r2, c2;    //分别是入口，方向，初始位置，终点位置
int has_edge[10][10][4][3];     //has_edge[r][c][dir][turn] = 1,表示在r,c位置，面向dir的点向turn转是可行的
int d[10][10][4];                       //d[r][c][dir] = m 表示在点r,c,面向dir方向的点距离初始状态的最近距离是m
Node parent[10][10][4];         //表示一个点的父节点

bool read_input() {
    char s[100], s1[100];
    if(scanf("%s%d%d%s%d%d", s, &r0, &c0, s1, &r2, &c2) != 6 )   return false;

    printf("%s\n", s);    //输出此次测试情况的名称

    dir = dirToInt(s1[0]);//求方向，将字母装化成数字
    r1 = r0 + dr[dir];
    c1 = c0 + dc[dir];  //求出初始状态

    memset(has_edge, 0, sizeof(has_edge));
    while(true) {
        int row, col;
        scanf("%d", &row);
        if(row == 0)    break;
        scanf("%d", &col);

        while(scanf("%s", s) == 1 && s[0] != '*') {
            int len = strlen(s);
            for(int i = 1; i < len; i++) {
                has_edge[row][col][dirToInt(s[0])][turnToInt(s[i])] = 1;  //读入has_edge数组
            }
        }
    }
    return true;
}

Node walk(const Node & u, int i) {
    int dir = u.dir;
    if(i == 1)  dir = (dir + 3) % 4;
    if(i == 2)  dir = (dir + 1) % 4;
    return Node(u.r + dr[dir], u.c + dc[dir], dir);
}

void print_ans(Node u) {
    vector<Node> vec;
    while(true) {
        vec.push_back(u);
        if(d[u.r][u.c][u.dir] == 0) break;
        u = parent[u.r][u.c][u.dir];
    }
    vec.push_back(Node(r0, c0, dir));

    int cnt = 0;
    int len = vec.size();
    for(int i = len - 1; i >= 0; i--) {
        if(cnt % 10 == 0)   printf(" ");
        printf(" (%d,%d)", vec[i].r, vec[i].c);
        if(++cnt % 10 == 0) printf("\n");
    }

    if(len % 10 != 0)    printf("\n");
}

bool inside(int r, int c) {
    return r >= 1 && r <= 9 && c >= 1 && c <= 9;
}

void solve() {
    queue<Node> q;

    Node u(r1, c1, dir);
    memset(d, -1, sizeof(d));
    d[u.r][u.c][u.dir] = 0;

    q.push(u);
    while(!q.empty()) {     //bfs算法
        Node u = q.front();     q.pop();
        if(u.r == r2 && u.c == c2) {
            print_ans(u);
            return;
        }

        for(int i = 0; i < 3; i++) {    //进入一个点后向三个方向试探
            Node v = walk(u, i);    //i = 0:直走 i = 1:左转 ；i = 2：右转
            if(has_edge[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;
                parent[v.r][v.c][v.dir] = u;
                q.push(v);
            }
        }
    }

    printf(" No Solution Possible\n");
}

int main()
{

// freopen("in.txt","r",stdin);
 // freopen("out.txt","w",stdout);

    while(read_input()) {
        solve();
    }
    return 0;
}