[uva816]AbbottsRevenge Abbott的复仇(经典迷宫BFS)

这题思路就普通的BFS加上一个维度朝向，主要是要注意输入，输出，以及细节的处理

 

#include<cstdio>

#include<cstring>

#include<queue>

#include<vector>

using namespace std;



const int MAX = 9+1;

const int DIR = 4;

const int TURN = 3;

bool have_edge[MAX][MAX][DIR][TURN];//所在位置 ， 朝向 ， 要走的方向



int d[MAX][MAX][DIR];//最短路径 同时可以判断结点是否访问过

#define Node(x,n) x[n.r][n.c][n.dir]



struct node

{

   int r , c, dir;

   node(){}

   node(int r, int c, int dir):r(r),c(c),dir(dir){}

};



struct node p[MAX][MAX][DIR];//父节点 打印 路径

const char *dirs = "NESW";//clockwise

const char *turns = "LFR";



inline int dir_id(char c) {return strchr(dirs,c)-dirs;}

inline int turn_id(char c) {return strchr(turns,c)-turns;}



const int MAX_NAME_LEN = 42;

char MazeName[MAX_NAME_LEN];//20个字符，还有空格。。。

int r0,c0;

int r1,c1,dir1;

int r2,c2;

int dr[] = {-1, 0, 1, 0};//clockwise

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



bool read()

{

   gets(MazeName);//lineread

   //  if(!strcmp(MazeName,"END")) return false;

   char s[10]; int r, c;

   if(!~scanf("%d%d%s",&r0,&c0,s)) return false;

   printf("%s\n",MazeName);

   memset(have_edge,false,sizeof(have_edge));



   dir1 = dir_id(s[0]);

   r1 = r0 + dr[dir1]; c1 = c0 + dc[dir1];

   scanf("%d%d",&r2,&c2);



   while(scanf("%d",&r),r) {

      scanf("%d",&c);

      while(scanf("%s",s),*s != '*'){

         char *p = s; int dir = dir_id(*s);//!!

         while(*(++p)) { have_edge[r][c][dir][turn_id(*p)] = true;}

      }

   }

   getchar();//吸收换行符

   return true;

}



void print_ans(node u)

{

   vector<node> ans;

   for(;;){

      ans.push_back(u);

      if(Node(d,u) == 0) break;

      u = Node(p,u);

   }

   ans.push_back(node(r0,c0,dir1));

   int cnt = 0;

   for(int i = ans.size() - 1; i >= 0; i--){

      if(cnt%10 == 0) printf(" ");

      printf(" (%d,%d)", ans[i].r, ans[i].c);

      if(++cnt%10 == 0) puts("");

   }

   if(ans.size()%10) puts("");//注意

}



node walk(node &u,int turn){

   int dir = u.dir;

   if(turn == 0) dir = (dir + 3)%4;//rev 

   else if(turn == 2) dir = (dir + 1)%4;//clockwise

   return node(u.r + dr[dir], u.c + dc[dir], dir );

}

inline bool inside(int r,int c) { return r > 0&&r <= 9 && c > 0 && c <= 9; }







void solve()

{

   queue<node> q;

   node u(r1,c1,dir1);//c1 - >c2

   memset(d, -1, sizeof(d));

   d[u.r][u.c][u.dir] = 0;

   q.push(u);

   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++) {

         if(!have_edge[u.r][u.c][u.dir][i]) continue;

         node v = walk(u,i);

         if(inside(v.r, v.c) && d[v.r][v.c][v.dir] < 0){

            d[v.r][v.c][v.dir] = Node(d,u) + 1;

            Node(p,v) = u;

            q.push(v);//push(u) ...

         }

      }

   }

   printf("  No Solution Possible\n");

}



int main()

{

#ifdef local

   freopen("in2.txt","r",stdin);

  // freopen("myout.txt","w",stdout);

#endif // local

   while(read()){

      solve();

   }

   return 0;

}