Abbott的复仇

题目：
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <queue>
 6 #include <vector>
 7 using namespace std;
 8 const int maxn = 10;
 9 struct Node{
10     int r;
11     int c;
12     int dir;
13     Node(int r = 0,int c = 0,int dir = 0):r(r),c(c),dir(dir){}
14 };
15 int has_edg[maxn][maxn][4][3]; //每个结点的转向信息 
16 int d[maxn][maxn][4];           //每个结点距离初始点的距离 
17 Node pre[maxn][maxn][4];
18 int r0,c0,r1,c1,r2,c2,dir;
19 const char* dirs = "NESW";        //从上右下左四个方向进入（顺时针） 
20 const char* turns = "FLR";        //直行、左拐、右拐四种转向 
21 int dir_id(char d){    return strchr(dirs,d) - dirs;}
22 int turn_id(char t){ return strchr(turns,t) - dirs;}
23 int dr[4] = {-1,0,1,0};   //NESW四个方向
24 int dc[4] = {0,1,0,-1};   
25 bool read_case(){
26     char s1[99],s2[99];
27     if(scanf("%d%d%s%d%d",&c0,&r0,s1,&r2,&c2) != 5) return false;
28     
29     dir = dir_id(s1[0]); //从给出的起点走一步，获得起始位置 
30     r1 = r0 + dr[dir];
31     c1 = c0 + dc[dir];
32     
33     memset(has_edg,0,sizeof(has_edg));
34     for(;;){
35         int c,r;
36         cin >> c;
37         if(c == 0) break;
38         cin >> r;
39         while(scanf("%s",s2) == 1 && s2[0] != '*'){
40             for(int i = 1;i < strlen(s2);i++){
41                 has_edg[c][r][dir_id(s2[0])][turn_id(s2[i])] = 1;
42             }
43         }     
44     }
45     return true; 
46 }  
47 Node walk(const Node& n,int turn){   //向前走一步 
48     int dir = n.dir; //turn == 0时直走，方向不变 
49     if(turn == 1) dir = (dir + 3) % 4; //左转(从进入时的方向顺时针数三个方向) 
50     if(turn == 2) dir = (dir + 1) % 4; //右转(从进入时的方向顺时针数一个方向) 
51     return Node(n.r + dr[dir],n.c + dc[dir],dir); 
52 } 
53 bool inside(int r,int c){
54     return r >= 1 && r <= 9 && c >= 1 && c <= 9;
55 }
56 
57 void print_ans(Node n){
58     vector<Node> nodes;
59     for(;;){
60         nodes.push_back(n);
61         if(d[n.r][n.c][n.dir] == 0) break;
62         n = pre[n.r][n.c][n.dir];
63     }
64     nodes.push_back(Node(r0,c0,dir));
65     int cnt = 0;
66     for(int i = nodes.size();i >= 0;i--){
67         cout << "(" << nodes[i].r << "," << nodes[i].c << ")" << " ";
68         cnt++;
69         if(cnt % 10 == 0) cout << "\n";
70     } 
71 }
72 void solve(){
73     queue<Node> q;
74     memset(d,-1,sizeof(d));
75     q.push(Node(r1,c1,dir));
76     d[r1][c1][dir] = 0; //初始点距离为0 
77     while(!q.empty()){
78         Node n = q.front(); q.pop();
79         if(n.r == r2 && n.c == c2){ print_ans(n); return;}
80         for(int i = 0;i < 3;i++){
81             if(has_edg[n.r][n.c][n.dir][i]){
82                 Node v = walk(n,i);    
83                 if(inside(v.r,v.c) && d[v.r][v.c][v.dir] == -1){
84                     d[v.r][v.c][v.dir] = d[n.r][n.c][n.dir] + 1;
85                     pre[v.r][v.c][v.dir] = n;
86                     q.push(v);
87                 }
88             }
89         }
90     }
91     printf("No Solution Possible\n");
92 }
93 int main(){
94     while(read_case()){
95         solve();
96     }
97     return 0;
98 }
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