回溯法( Backtracking Algorithms ) ：C語言Maze迷宮問題（自己實現）

http://www.cs.rpi.edu/~hollingd/psics/notes/backtracking.pdf

 Two situations:

– Finding a solution to a problem can't be based on a straight path to the goal.

● consider traversing a maze.

– We need a better approach than brute force(independently evaluating all possible solutions).

● Think of the TSP problem – many possible solutions sharepartial tours (why not treat identical partial tours as a singlepartial solution?)

TSP:旅行推銷員問題

http://en.wikipedia.org/wiki/Backtracking

http://baike.baidu.com/view/45.htm

自己實現的迷宮問題：

#include <stdio.h>#include <stdlib.h>#define true 1#define false 0typedef char byte;typedef struct OneCell{    byte up;    byte down;    byte left;    byte right;    int step;} Cell;typedef struct Pos{    int x;    int y;}Pos;int Maze(Cell *pArr,Pos * pCurr, Pos * pDest, Pos * pSize ,int step){Cell *pC = pArr + pCurr->x * pSize->y + pCurr->y ;    int cRow=pCurr->x;    int cCol=pCurr->y;    int eRow=pDest->x;    int eCol=pDest->y;    pC->step=step;  //第幾步    if(cRow==eRow && cCol==eCol){        int i=0,j=0;        printf("\nSuccess match!\n");        for(;i<pSize->x;++i){            for(j=0;j<pSize->y;++j){                printf("%4d",pArr[i*pSize->y+j].step);            }            printf("\n");        }        printf("Step=%d:[%d,%d]\n",pC->step,cRow,cCol);        return true;    }    if(pC->right==true && cCol<(pSize->y-1) && pArr[cRow*pSize->y + cCol+1].step == -1 ){        Pos pCurrNew;        pCurrNew.x=cRow;        pCurrNew.y=cCol+1;        if(Maze(pArr,&pCurrNew,pDest,pSize,step+1)==true){            printf("Step=%d:[%d,%d]\n",pC->step,cRow,cCol);            return true;        }    }    if(pC->down==true && cRow <(pSize->x-1) && pArr[(cRow+1)*pSize->y + cCol].step == -1){        Pos pCurrNew;        pCurrNew.x=cRow+1;        pCurrNew.y=cCol;        if(Maze(pArr,&pCurrNew,pDest,pSize,step+1)==true){            printf("Step=%d:[%d,%d]\n",pC->step,cRow,cCol);            return true;        }    }    if(pC->left==true && cCol>0 && pArr[cRow*pSize->y+cCol-1].step == -1 ){        Pos pCurrNew;        pCurrNew.x=cRow;        pCurrNew.y=cCol-1;        if(Maze(pArr,&pCurrNew,pDest,pSize,step+1)==true){            printf("Step=%d:[%d,%d]\n",pC->step,cRow,cCol);            return true;        }    }    if(pC->up==true && cRow>0 && pArr[(cRow-1)*pSize->y+cCol].step == -1 ){        Pos pCurrNew;        pCurrNew.x=cRow-1;        pCurrNew.y=cCol;        if(Maze(pArr,&pCurrNew,pDest,pSize, step+1)==true){            printf("Step=%d:[%d,%d]\n",pC->step,cRow,cCol);            return true;        }    }    pC->step=-1;    return false;}int main(){    Cell cells[][4]=    {        {            {false,true,false, false,-1},            {false,true, false, false,-1},            {false,true, false, true,-1},            {false,false,true, false,-1}        },        {            {true,true,false,true,-1},            {true,false,true,false,-1},            {true, true,false,true,-1},            {false,true, true,false,-1}        },        {            {true ,false ,false ,true,-1},            {false,false,true ,true,-1},            {true ,false ,true,false,-1},            {true ,false ,true,true,-1}        }    };    //cells[0][0].step=0;    Pos pCurr={0,0},pDest={2,3},pSize={3,4}; // rows, cols    Maze((Cell *)cells,&pCurr,&pDest,&pSize,0 );    //true;    //printf("%d,%d,%d ",sizeof (Cell),false,c1.down );    return 0;}

　　結果：

Success match!   0  -1  -1  -1   1  -1   5   6   2   3   4   7Step=7:[2,3]Step=6:[1,3]Step=5:[1,2]Step=4:[2,2]Step=3:[2,1]Step=2:[2,0]Step=1:[1,0]Step=0:[0,0]Process returned 0 (0x0)   execution time : 0.016 sPress any key to continue.