優先隊列 lag pri HERE out space sta quick diag

Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

Is an escape possible? If yes, how long will it take?

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a ‘#‘ and empty cells are represented by a ‘.‘. Your starting position is indicated by ‘S‘ and the exit by the letter ‘E‘. There‘s a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line

Trapped!

簡單的說就是在三維地圖中找最短路，具體題目請參見POJ2251

一、普通的bfs,配合優先隊列

代碼如下：

  1 #include<stdio.h>
  2 #include<iostream>
  3 #include<queue>
  4 using namespace std;
  5 char map[30][30][30];        //記錄節點信息
  6 int sta[30][30][30];        //標記是否訪問
  7 int base[6][3] = { {-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1} };
  8 int L, R, C;
  9 struct Piont
 10 {
 11     int x, y, z;            //位置坐標
 12     int step;                //出發點到該點的步數
 13 };
 14 struct Piont s;                //起點
 15 struct Piont e;                //終點
 16 struct Piont curp;            //跳出循環時的節點
 17 
 18 /******************判斷是否到達終點*********************/
 19 bool success(struct Piont cur)
 20 {
 21     if (cur.x == e.x && cur.y == e.y && cur.z == e.z)
 22         return true;
 23     else
 24         return false;
 25 }
 26 
 27 /**************判斷該點是否合法*************************/
 28 bool check(int x, int y, int z)
 29 {
 30     if ((x >= 0) && (x < L) && (y >= 0) && (y < R) && (z >= 0) && (z < C) && (!sta[x][y][z]) && (map[x][y][z] == ‘.‘ || map[x][y][z] == ‘E‘))
 31         return true;
 32     else
 33         return false;
 34 }
 35 
 36 /*************************深搜***************************/
 37 void bfs()
 38 {
 39     struct Piont next;
 40     queue<Piont>q;
 41     q.push(s);
 42     //int flag = 0;
 43     while (!q.empty())
 44     {
 45         curp = q.front();
 46         q.pop();
 47         if (success(curp))
 48             return;
 49         else
 50         {
 51             sta[curp.x][curp.y][curp.z] = 1;
 52             for (int i = 0; i < 6; i++)
 53             {
 54                 next.x = curp.x + base[i][0];
 55                 next.y = curp.y + base[i][1];
 56                 next.z = curp.z + base[i][2];
 57                 if (check(next.x, next.y, next.z))        //擴展隊列
 58                 {
 59                     next.step = curp.step + 1;
 60                     sta[next.x][next.y][next.z] = 1;
 61                     q.push(next);
 62                 }
 63              }
 64         }
 65     }
 66 }
 67 int main()
 68 {
 69     while (scanf("%d%d%d", &L, &R, &C))
 70     {
 71         if((L == 0) && (R == 0) && (C == 0))
 72             break;
 73         memset(sta, 0, sizeof(sta));
 74         for (int i = 0; i < L; i++) {
 75             getchar();
 76             for (int j = 0; j < R; j++) {
 77                 for (int k = 0; k < C; k++)
 78                 {
 79                     scanf("%c", &map[i][j][k]);
 80                     if (map[i][j][k] == ‘S‘) {
 81                         s.x = i;
 82                         s.y = j;
 83                         s.z = k;
 84                         s.step = 0;
 85                     }
 86                     else if (map[i][j][k] == ‘E‘)
 87                     {
 88                         e.x = i;
 89                         e.y = j;
 90                         e.z = k;
 91                     }
 92                 }
 93                 getchar();
 94             }
 95         }
 96         bfs();
 97         if (curp.x == e.x && curp.y == e.y && curp.z == e.z)
 98             printf("Escaped in %d minute(s).\n", curp.step);
 99         else
100             printf("Trapped!\n");
101     }
102     return 0;
103 }

二、遞歸（但由於多次重復經過某點，時間復雜度遠大於方法一）

僅供參考，代碼如下：

  1 #include<stdio.h>
  2 #include<iostream>
  3 using namespace std;
  4 char map[30][30][30];
  5 int step_map[30][30][30];
  6 int sta[30][30][30];
  7 int s_x = -1, s_y = -1, s_z = -1;
  8 int e_x = -1, e_y = -1, e_z = -1;
  9 int step = 0, minn = 1 << 25;
 10 int L, R, C;
 11 int base[6][3] = { {-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1} };
 12 
 13 bool check(int x, int y, int z)
 14 {
 15     if ((x >= 0) && (x < L) && (y >= 0) && (y < R) && (z >= 0) && (z < C))
 16         return true;
 17     else
 18         return false;
 19 }
 20 void bfs(int x, int y, int z)
 21 {
 22     int temp_x, temp_y, temp_z;
 23     for (int i = 0; i < 6; i++)
 24     {
 25         if (x == e_x + base[i][0] && y == e_y + base[i][1] && z == e_z + base[i][2])
 26         {
 27             if (step < minn)
 28                 minn = step;
 29             return;
 30         }
 31     }
 32     for (int i = 0; i < 6; i++)
 33     {
 34         temp_x = x + base[i][0];
 35         temp_y = y + base[i][1];
 36         temp_z = z + base[i][2];
 37         if ((!sta[temp_x][temp_y][temp_z]) && (map[temp_x][temp_y][temp_z] == ‘.‘) && (check(temp_x, temp_y, temp_z)))
 38         {
 39             step++;
 40             if (step < step_map[temp_x][temp_y][temp_z])        //剪枝二：當前步數已大於曾經過該點的最小步數，停止搜索
 41             {
 42                 step_map[temp_x][temp_y][temp_z] = step;        
 43                 if (step < minn)                                //剪枝一：當前步數已大於或等於最小步數，停止搜索
 44                 {
 45                     sta[temp_x][temp_y][temp_z] = 1;
 46                     bfs(temp_x, temp_y, temp_z);
 47                     sta[temp_x][temp_y][temp_z] = 0;
 48                 }
 49             }
 50             step--;
 51         }
 52     }
 53 }
 54 int main()
 55 {
 56     while (scanf("%d%d%d",&L,&R,&C))
 57     {
 58         if ((L == 0) && (R == 0) && (C == 0))
 59             break;
 60         memset(sta, 0, sizeof(sta));
 61         //memset(step_map, (1 << 25), sizeof(step_map));//只能用來初始化為0、1和-1
 62         for (int i = 0; i < 30; i++)
 63             for (int j = 0; j < 30; j++)
 64                 for (int k = 0; k < 30; k++)
 65                     step_map[i][j][k] = (1 << 25);
 66         
 67         for (int i = 0; i < L; i++) {
 68             getchar();
 69             for (int j = 0; j < R; j++) {
 70                 for (int k = 0; k < C; k++)
 71                 {
 72                     //cin >> map[i][j][k];
 73                     scanf("%c", &map[i][j][k]);
 74                     if (map[i][j][k] == ‘S‘) {
 75                         s_x = i;
 76                         s_y = j;
 77                         s_z = k;
 78                     }
 79                     if (map[i][j][k] == ‘E‘)
 80                     {
 81                         e_x = i;
 82                         e_y = j;
 83                         e_z = k;
 84                     }
 85                 }
 86                 getchar();
 87             }        
 88         }
 89                 
 90         bfs(s_x, s_y, s_z);
 91         if (minn == (1 << 25))
 92             printf("Trapped!");
 93         else
 94         {
 95             printf("Escaped in %d minnute(s).", minn + 1);
 96             minn = (1 << 25);
 97             step = 0;
 98         }
 99     }
100     return 0;
101 }

新手入門，希望大家多多指教！

Dungeon Master的兩種方法