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迷宫求解 《数据结构——严蔚敏》

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推荐:数据结构之栈的应用----迷宫求解

/***********程序设计思想*************/ (1)迷宫地图相关: 利用动态二维数组来初步勾勒出迷宫: 建议先用malloc申请一维数组,再用calloc申请每个元素中的一

  在学《数据结构——严蔚敏》,自己动手写了 迷宫求解 的实现!为了方便,C和C++部分混写了= =!稍稍修改了点地方

 

推荐:数据结构:栈应用_求解迷宫问题3

/************************************************************************/ /* 数据结构:栈应用:求解迷宫问题                                          

 

/********************************************** *Name: 迷宫求解 *Date: 2010-09-25 *Author: Kleist *From:《数据结构 ——严蔚敏》 ***********************************************/ #define OK 1 #define ERROR 0 #define TRUE 1 #define FALSE 0 #define OVERFLOW -2 #define STACK_INIT_SIZE 100 #define STACKINCREMENT 10 typedef int Status; typedef struct { int x; int y; }PosType; typedef struct { int ord; //通道块在路径上的“序号” PosType seat; //通道块在迷宫中的“坐标位置” int di; //从此通道块走向下一通道块的“方向”, //1:东 2:北 3:西 (顺时针) }MazeType; typedef struct { MazeType *base; MazeType *top; int stacksize; }MazeStack; #include <iostream> using namespace std; Status InitStack(MazeStack &S); Status Push(MazeStack &S, MazeType &e); Status Pop(MazeStack &S, MazeType &e); Status StackEmpty(MazeStack &S); Status MazePath(PosType start, PosType end); Status Pass(PosType &pos); void FootPrint(PosType pos); PosType NextPos(PosType curPos, int &i); void MakePrint(PosType pos); //迷宫地图,0表示墙壁,1表示通路,入口:mazeMap[1][1],出口mazeMap[8][8] int mazeMap[10][10] = { //0,1,2,3,4,5,6,7,8,9 {0,0,0,0,0,0,0,0,0,0}, //0 {0,1,1,0,1,1,1,0,1,0}, //1 {0,1,1,0,1,1,1,0,1,0}, //2 {0,1,1,1,1,0,0,1,1,0}, //3 {0,1,0,0,0,1,1,1,1,0}, //4 {0,1,1,1,0,1,1,1,1,0}, //5 {0,1,0,1,1,1,0,1,1,0}, //6 {0,1,0,0,0,1,0,0,1,0}, //7 {0,0,1,1,1,1,1,1,1,0}, //8 {0,0,0,0,0,0,0,0,0,0} //9 }; int main() { PosType mazeStart, mazeEnd; mazeStart.x = 1; mazeStart.y = 1; mazeEnd.x = 8; mazeEnd.y = 8; cout << "迷宫: " << endl; for(int i = 0; i < 10; ++i) { for(int j = 0; j < 10; ++j) cout << mazeMap[i][j]; cout << endl; } cout << endl << endl; if(MazePath(mazeStart, mazeEnd)) cout << "/n走通迷宫" << endl; else cout << "/n走不通迷宫" << endl; system("PAUSE"); return 0; } Status InitStack(MazeStack &S) { S.base = (MazeType *)malloc(STACK_INIT_SIZE * sizeof(MazeType)); if(!S.base) exit(OVERFLOW); S.top = S.base; S.stacksize = STACK_INIT_SIZE; return OK; } Status Push(MazeStack &S, MazeType &e) { if(S.top - S.base >= S.stacksize) { S.base = (MazeType *)realloc(S.base, (S.stacksize + STACKINCREMENT) * sizeof(MazeType)); if(!S.base) exit(OVERFLOW); S.top = S.base + S.stacksize; S.stacksize += STACKINCREMENT; } *S.top++ = e; return OK; } Status Pop(MazeStack &S, MazeType &e) { if(S.top == S.base) return ERROR; e = *--S.top; return OK; } Status StackEmpty(MazeStack &S) { if(S.base == S.top) return OK; return ERROR; } Status MazePath(PosType start, PosType end) { PosType curpos; MazeStack S; MazeType e; int curstep; InitStack(S); curpos = start; //设定当前位置为入口位置 curstep = 1; //探索第一步 cout << "起点: "<< "(" <<start.y << "," << start.x << ")" << endl; do { if(Pass(curpos)) //当前位置可以通过,即是未曾走到的通道块 { FootPrint(curpos); //留下足迹 e.ord = curstep; e.seat = curpos; e.di = 1; Push(S, e); //加入路径 if(curpos.x == end.x && curpos.y == end.y) { cout << "/n终点 (" << e.seat.y << "," << e.seat.x << ")"; return TRUE; //到达终点(出口) } curpos = NextPos(curpos, e.di); //下一位置是当前位置的东邻 ++curstep; //探索下一步 } else //当前位置不能通过 { if(!StackEmpty(S)) { Pop(S, e); while(e.di == 4 && !StackEmpty(S)) { MakePrint(e.seat); //留下不能通过的标记 Pop(S, e); cout << "倒退到(" << e.seat.y << "," << e.seat.x << ")"; } if(e.di < 4) { ++e.di; //换下一个方向探索 Push(S, e); curpos = NextPos(e.seat, e.di); //设定当前位置是该新方向上的相邻块 } } } }while(!StackEmpty(S)); return FALSE; } Status Pass(PosType &pos) { if(mazeMap[pos.y][pos.x] == 0) return FALSE; cout << "->(" << pos.y << "," << pos.x << ")"; return TRUE; } void FootPrint(PosType pos) { mazeMap[pos.y][pos.x] = 2; //将走过的路径设为2 } PosType NextPos(PosType curPos, int &i) { switch(i) //顺时针方向 { case 1: ++curPos.x; //东 if(mazeMap[curPos.y][curPos.x] != 2) break; --curPos.x; case 2: i = 2; ++curPos.y; //南 if(mazeMap[curPos.y][curPos.x] != 2) break; --curPos.y; case 3: i = 3; --curPos.x; //西 if(mazeMap[curPos.y][curPos.x] != 2) break; ++curPos.x; case 4: i = 4; --curPos.y; //北 if(mazeMap[curPos.y][curPos.x] == 2) { ++curPos.y; mazeMap[curPos.y][curPos.x] = 0; } break; } return curPos; } void MakePrint(PosType pos) { cout << "/n(" << pos.y << "," << pos.x << ")走不通,作废"; mazeMap[pos.y][pos.x] = 0; //将走不通的块替换为墙壁 }

推荐:数据结构(迷宫求解c++)

#include "sqstack.h"Status InitStack(Stack &S){ S.top = 0 ; for( int i = 0 ; i < MaxSize ; i++ ) { S.data[i].di = 0 ; S.data[i]

  在学《数据结构——严蔚敏》,自己动手写了 迷宫求解 的实现!为了方便,C和C++部分混写了= =!稍稍修改了点地方     /********************************************** *Name: 迷宫求解 *Dat

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