迷宫最短路径（BFS）

　　用队列实现广度优先搜索（BFS），找出最短路径。用栈保存走过的路径，并输出路径和标识最短路径的地图。

输入用例： 0：路 1：墙壁

24 24
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0
0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0
0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0
0 1 1 1 0 1 1 0 0 1 1 1 0 1 1 0 0 1 1 1 0 1 1 0
1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0
0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0
0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0
0 1 1 1 0 1 1 0 0 1 1 1 0 1 1 0 0 1 1 1 0 1 1 0
1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0
0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0
0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0
0 1 1 1 0 1 1 0 0 1 1 1 0 1 1 0 0 1 1 1 0 1 1 0
1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0

　　这里需要在头文件中重新定义队列和栈的数据元素类型 SElemType, QElemType。

define.h

// define.h
#ifndef __MENGQL_DEFINE__
#define __MENGQL_DEFINE__

#define C_LOG_DBG(format, ...) 
//printf("[%s@%s,%d] " format ,__FUNCTION__, __FILE__, __LINE__, ##__VA_ARGS__);
#define C_LOG_ERR(format, ...) printf("[%s@%s,%d] " format ,__FUNCTION__, __FILE__, __LINE__, ##__VA_ARGS__);
typedef enum EStatus {ERROR, OK} Status;

typedef struct
{
    int x;
    int y;
}PosType;

typedef struct
{
    PosType seat;
    PosType parent;
}SElemType, QElemType;

#endif

Maze.c

  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #include "define.h"
  4 #include "SqQueue.h"
  5 #include "SqStack.h"
  6 
  7 int **g_MazeMap;
  8 int g_m, g_n;//g_m:列数 g_n:行数
  9 
 10 #define SHORTPRINT 5
 11 #define FOOTPRINT 2
 12 
 13 void MazePrint()
 14 {
 15     int i, j;
 16     for(i=0; i<g_m; ++i)
 17     {
 18         for(j=0; j<g_n; ++j)
 19         {
 20             printf("%d ", g_MazeMap[i][j]);
 21         }
 22         printf("\n");
 23     }
 24 }
 25 void MarkShortPrint(PosType pos)
 26 {
 27     g_MazeMap[pos.y][pos.x] = SHORTPRINT;
 28 }
 29 void MazePrintRes(SqStack *S, int x, int y)
 30 {
 31     SElemType e;
 32     while(StackEmpty(S) != OK)
 33     {
 34         PopStack(S, &e);
 35         if(e.seat.x == x && e.seat.y == y)
 36         {
 37             MazePrintRes(S, e.parent.x, e.parent.y);
 38             MarkShortPrint(e.seat);
 39             printf("[%d][%d]\n", x, y);
 40         }
 41     }
 42     return;
 43 }
 44 Status MazePass(PosType pos)
 45 {
 46     if(pos.x>=0 && pos.y>=0 && pos.x<g_m && pos.y<g_n && g_MazeMap[pos.y][pos.x] == 0)
 47     {
 48         return OK;
 49     }
 50     return ERROR;
 51 }
 52 void MazeFootPrint(PosType pos)
 53 {
 54     g_MazeMap[pos.y][pos.x] = FOOTPRINT;
 55 }
 56 void NextPos(PosType *pos, int di)
 57 {
 58     switch(di)
 59     {
 60         case 1:
 61             pos->y = pos->y-1;
 62             break;
 63         case 2:
 64             pos->x = pos->x-1;
 65             break;
 66         case 3:
 68             pos->y = pos->y+1;
 69             break;
 70         case 4:
 71             pos->x = pos->x+1;
 72             break;
 73         defult:
 74             break;
 75     }
 76 }
 77 Status MazeShortPath(PosType *start, PosType *end)
 78 {
 79     PosType curPos, nextPos;
 80     SElemType e;
 81     SqQueue Q;
 82     SqStack resS;//保存遍历过的节点，用于输出结果
 83     int i;
 84     
 85     e.parent.x = -1;
 86     e.parent.y = -1;
 87     e.seat.x = start->x;
 88     e.seat.y = start->y;
 89     
 90     if(InitQueue(&Q) != OK || InitStack(&resS) != OK)
 91     {
 92         return ERROR;
 93     }
 94     do
 95     {
 96         PushStack(&resS, e);
 97         curPos.x = e.seat.x;
 98         curPos.y = e.seat.y;
 99         for(i=1; i<5; ++i)
100         {
101             C_LOG_DBG("[%d][%d][%d]\n", curPos.x, curPos.y, i);
102             nextPos.x = curPos.x;
103             nextPos.y = curPos.y;
104             NextPos(&nextPos, i);
105             if(MazePass(nextPos) == OK)
106             {
107                 C_LOG_DBG("[%d][%d][%d][%d][%d]\n", curPos.x, curPos.y, nextPos.x, nextPos.y);
108                 e.parent.x = curPos.x;
109                 e.parent.y = curPos.y;
110                 e.seat.x = nextPos.x;
111                 e.seat.y = nextPos.y;
112                 
113                 EnQueue(&Q, e);
114                 MazeFootPrint(nextPos);
115                 
116                 if(nextPos.x == end->x && nextPos.y == end->y)
117                 {
118                     C_LOG_DBG("%s\n", "DEBUG");
119                     PushStack(&resS, e);
120                     MazePrintRes(&resS, nextPos.x, nextPos.y);
121                     MazePrint();
122                     return OK;
123                 }
124             }
125         }
126     }while(DeQueue(&Q, &e) == OK);
127     MazePrint();
128     DestoryStack(&resS);
129     DestroyQueue(&Q);
130     return ERROR;
131 }
132 
133 int main()
134 {
135     int i, j;
136     PosType start;
137     PosType end;
138     
139     scanf("%d %d", &g_m, &g_n);
140     start.x = 0;
141     start.y = 0;
142     end.x = g_m-1;
143     end.y = g_n-1;
144     
145     g_MazeMap=(int**)malloc(g_m*sizeof(int*));
146     for(i=0; i<g_m; ++i)
147     {
148         g_MazeMap[i] = (int*)malloc(g_n*sizeof(int));
149     }
150     for(i=0; i<g_m; ++i)
151     {
152         for(j=0; j<g_n; ++j)
153         {
154             scanf("%d", &g_MazeMap[i][j]);
155         }
156     }
157     MazeShortPath(&start, &end);
158 }

输出结果（路径坐标）：

[0][0]
[0][1]
[0][2]
[0][3]
[0][4]
[1][4]
[2][4]
[2][5]
[3][5]
[4][5]
[4][6]
[4][7]
[5][7]
[6][7]
[7][7]
[7][8]
[7][9]
[7][10]
[7][11]
[7][12]
[8][12]
[9][12]
[10][12]
[10][13]
[11][13]
[12][13]
[12][14]
[12][15]
[13][15]
[14][15]
[15][15]
[15][16]
[15][17]
[15][18]
[15][19]
[15][20]
[16][20]
[17][20]
[18][20]
[18][21]
[19][21]
[20][21]
[20][22]
[20][23]
[21][23]
[22][23]
[23][23]

输出结果（路径地图）：0：没走过的路 1：墙壁 2：走过的路 5：走过的路中的最短路径

5 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 
5 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 
5 2 2 2 1 1 2 2 2 2 2 2 1 1 2 2 2 2 2 2 1 1 2 2 
5 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 
5 5 5 1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 2 
2 1 5 5 5 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 
2 1 1 1 5 1 1 2 2 1 1 1 2 1 1 2 2 1 1 1 2 1 1 2 
1 2 2 2 5 5 5 5 1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 
2 2 1 2 2 2 1 5 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 
2 2 1 2 2 2 1 5 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 
2 2 2 2 1 1 2 5 2 2 2 2 1 1 2 2 2 2 2 2 1 1 2 2 
2 1 1 1 1 2 2 5 2 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 
2 2 2 1 2 2 2 5 5 5 5 1 2 2 2 2 2 2 2 1 2 2 2 2 
2 1 2 2 2 1 2 2 2 1 5 5 5 1 2 2 2 1 2 2 2 1 2 2 
2 1 1 1 2 1 1 2 2 1 1 1 5 1 1 2 2 1 1 1 2 1 1 2 
1 2 2 2 2 2 2 2 1 2 2 2 5 5 5 5 1 2 2 2 2 2 2 2 
2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 5 2 2 1 2 2 2 1 2 
2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 5 2 2 1 2 2 2 1 2 
2 2 2 2 1 1 2 2 2 2 2 2 1 1 2 5 2 2 2 2 1 1 2 2 
2 1 1 1 1 2 2 2 2 1 1 1 1 2 2 5 2 1 1 1 1 2 2 2 
2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 5 5 5 5 1 2 2 2 2 
2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 5 5 5 1 2 2 
2 1 1 1 2 1 1 2 2 1 1 1 2 1 1 2 2 1 1 1 5 1 1 0 
1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 5 5 5 5

 

posted on 2012-08-22 16:42  favourmeng  阅读(984)  评论(0)  编辑  收藏  举报

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