Poj 3083 Children of the Candy Corn

  1 #include <iostream>
  2 #include <queue>
  3 #define N 50
  4 const int west = 0, north = 1, east = 2, south = 3;
  5 const int left = 0, forw = 1, right = 2, back = 3;
  6 char m[N][N];
  7 int n, w, h, x, y;
  8 int d[4][2][4] = { //d[i][j][k]代表从当前点朝k方向时往i方向排列应改变的值
  9 { { 1, 0, -1, 0 },{ 0, -1, 0, 1 } },
 10 { {0, -1, 0, 1}, { -1, 0, 1, 0 } },
 11 { {-1, 0, 1, 0}, { 0, 1, 0, -1 } },
 12 { {0, 1, 0, -1}, { 1, 0, -1, 0 } }
 13 };
 14 int dirc[4] = {3,0,1,2};
 15 void solve1(int sr,int sc,int sdir){
 16     int res=2,r=sr,c=sc,dir=sdir;
 17     while (m[r][c] != ‘E‘){
 18         for (int i = 0; i < 4; i++){//当前位置方向下，依次朝左右前后四个方向尝试移动，即左优先
 19             if (m[r + d[i][0][dir]][c + d[i][1][dir]] != ‘#‘){
 20                 r += d[i][0][dir];
 21                 c += d[i][1][dir];
 22                 dir = (dir+dirc[i])%4;
 23                 res++;
 24                 break;
 25             }
 26         }
 27     }
 28     printf("%d ", res);
 29     res = 2, r = sr, c = sc,dir=sdir;
 30     while (m[r][c] != ‘E‘){//右优先，因为对方向的定义，故不能像上面那样直接枚举
 31         int i;
 32         for (i = 2; i >=0; i--){
 33                 if (m[r + d[i][0][dir]][c + d[i][1][dir]] != ‘#‘){
 34                     r += d[i][0][dir];
 35                     c += d[i][1][dir];
 36                     dir = (dir+dirc[i])%4;
 37                     res++;
 38                     break;
 39                 }
 40         }
 41         if (i<0)
 42             if (m[r + d[back][0][dir]][c + d[back][1][dir]] != ‘#‘){
 43                     r += d[back][0][dir];
 44                     c += d[back][1][dir];
 45                     dir = (dir+dirc[back])%4;
 46                     res++;
 47             }
 48     }
 49     printf("%d ", res);
 50 }
 51 struct loca
 52 {
 53     int r, c, d,step;//定义节点为当前结点位置，方向及步数
 54 };
 55 
 56 int bfs(int sr,int sc,int sdir){
 57     queue<loca> q;
 58     loca p = { sr, sc, sdir,2},t;
 59     q.push(p);
 60     while (!q.empty())
 61     {
 62         p = q.front(); q.pop();
 63         for (int i = 0; i < 3; i++){
 64             if (m[p.r + d[i][0][p.d]][p.c + d[i][1][p.d]] != ‘#‘)
 65             {
 66                 if (m[p.r + d[i][0][p.d]][p.c + d[i][1][p.d]] == ‘E‘)return p.step+1;
 67                 m[p.r + d[i][0][p.d]][p.c + d[i][1][p.d]] = ‘#‘;
 68                 t.r = p.r + d[i][0][p.d];
 69                 t.c = p.c + d[i][1][p.d];
 70                 t.d = (p.d + dirc[i]) % 4;
 71                 t.step = p.step + 1;
 72                 q.push(t);
 73             }
 74         }
 75     }
 76     return 0;
 77 }
 78 void work(){
 79     int dir;
 80     for (dir = 0; dir < 4; dir++)
 81         if (m[y + d[forw][0][dir] ][x + d[forw][1][dir]] != ‘#‘){
 82             y += d[forw][0][dir];
 83             x += d[forw][1][dir];
 84             break;
 85         }
 86     solve1(y, x, dir);
 87     printf("%d\n",bfs(y,x,dir));
 88 }
 89 int main()
 90 {
 91     scanf("%d", &n);
 92     while (n--){
 93         scanf("%d%d", &w, &h);
 94         for (int i = 1; i <= h; i++){
 95             scanf("%s", m[i] + 1);
 96             m[i][0] = m[i][w+1] = ‘#‘;
 97         }//初始化将边界也变为#，减小编程难度
 98         for (int i = 0; i <= w+1; i++)
 99             m[0][i] = m[h+1][i] = ‘#‘;
100         for (int i = 1; i <= h; i++)
101             for (int j = 1; j <= w; j++){
102                 if (m[i][j] == ‘S‘){
103                     y = i;
104                     x = j;
105                 }
106         }
107         work();
108     }
109     return 0;
110 }
111