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(980) 评论(0) 编辑 收藏 举报