HDU 2102 A计划 _BFS

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #define CL(x,y) memset(x,y,sizeof(x))
  6 using namespace std;
  7 struct node
  8 {
  9     int x;
 10     int y;
 11     int z;
 12     int time;
 13 };
 14 queue<node> Q;
 15 const int MAX = 13;
 16 int cases, N, M, T;
 17 char maze[2][MAX][MAX];
 18 int used[2][MAX][MAX];
 19 bool checked(int y, int z);
 20 void BFS();
 21 int Move[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
 22 int main()
 23 {
 24     int i, j, k;
 25     cin >> cases;
 26     while(cases--)
 27     {
 28         cin >> N >> M >> T;
 29         getchar();
 30         for(k = 0; k < 2; k++)
 31         {
 32             for(i = 0; i < N; i++)
 33             {
 34                 for(j = 0; j < M; j++)
 35                     cin >> maze[k][i][j];
 36                 getchar();
 37             }
 38             if(k == 0)
 39                 getchar();
 40         }
 41         CL(used, 0);
 42         BFS();
 43         while(!Q.empty())
 44             Q.pop();
 45     }
 46     return 0;
 47 }
 48 void BFS()
 49 {
 50     int xx, yy, zz, i;
 51     node front, rear, cur;
 52     front.x = 0;
 53     front.y = 0;
 54     front.z = 0;
 55     front.time = 0;
 56     used[front.x][front.y][front.z] = 1;
 57     Q.push(front);
 58     while(!Q.empty())
 59     {
 60         cur = Q.front();
 61         Q.pop();
 62         if(cur.time > T)
 63         {
 64             cout << "NO" << endl;
 65             return ;
 66         }
 67         if(maze[cur.x][cur.y][cur.z] == ‘P‘)
 68         {
 69 //            cout << cur.time << endl;
 70             cout << "YES" << endl;
 71             return ;
 72         }
 73         for(i=0; i<4; i++)
 74         {
 75             xx = cur.x;
 76             yy = Move[i][0]+cur.y;
 77             zz = Move[i][1]+cur.z;
 78             if(checked(yy,zz) && maze[xx][yy][zz]!=‘*‘ && maze[xx][yy][zz]!=‘#‘ && !used[xx][yy][zz])
 79             {
 80                 used[xx][yy][zz] = 1;
 81                 rear.x = xx;//cur.x;这里的坐标一致都没发现,66666，WA到现在
 82                 rear.y = yy;//cur.y;
 83                 rear.z = zz;//cur.z;
 84                 rear.time = cur.time+1;
 85                 Q.push(rear);
 86             }
 87             if(checked(yy,zz) && maze[xx][yy][zz]==‘#‘ && !used[xx][yy][zz])//两个if()语句的顺序还是有一定规范的
 88             {
 89                 used[xx][yy][zz] = 1;
 90                 xx = (cur.x==1) ? 0 : 1;//如果下面的if()语句不执行，则只是改变了所在的层数而已
 91                 if(maze[xx][yy][zz]==‘*‘ || maze[xx][yy][zz]==‘#‘)//该路不可走
 92                 {
 93                     xx = (cur.x==1) ? 0 : 1;//变回原来的样子
 94                     used[xx][yy][zz] = 1;
 95                 }
 96                 else
 97                 {
 98                     rear.x = xx;
 99                     rear.y = yy;
100                     rear.z = zz;
101                     rear.time = cur.time+1;
102                     Q.push(rear);
103                 }
104             }
105         }
106     }
107     cout << "NO" << endl;
108     return ;
109 }
110 bool checked(int y, int z)
111 {
112     if(y>=0 && y<N && z>=0 && z<M)
113         return true;
114     return false;
115 }

两层，数组选择很重要，然后是对于结构体队列的运用，新建node front、rear、cur，进行条件的判断，两个if()语句的顺序不想考虑了，太烦