经典c程序(0036) ---迷宫求解(单条通路 栈实现)

/************************************************************************************** 
* Function     : test
* Create Date  : 2014/07/13
* Author       : NTSK13 
* Email        : beijiwei@qq.com 
* Copyright    : 欢迎大家和我一起交流学习，转载请保持源文件的完整性。 
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* Version      : V0.1                    
***************************************************************************************
经典c程序(0036) ---迷宫求解(单条通路 栈实现)

题目：较复杂迷宫求解   有多个分叉口,但只有一条路径  .

0为墙,1为通道

start (0,0)  end (9,9)
入口:

1 0 0 1 0 1 0 0 0 0
1 1 0 1 0 1 0 0 0 0
0 1 1 1 1 1 1 0 0 0
1 0 0 0 0 1 0 0 0 0
1 1 1 1 1 1 1 0 0 0
0 0 1 0 0 0 0 0 0 0
1 0 1 0 1 1 1 1 1 1
1 1 1 0 1 0 0 1 0 0
0 0 1 1 1 0 1 1 1 0
0 0 1 0 1 1 0 0 1 1 出口
     
**************************************************************************************/  
#include<stdio.h>

#define M 10
int data[M][M];

/***********************************************************/
typedef struct {
	int x;
	int y;
	int len;
}Pos;

Pos start,end;

Pos cross_store[M*M];
int cross_increase;

Pos path_store[M*M];
int path_increase;

int len;
/***********************************************************/
int offset[4][2]={-1,0,0,1,1,0,0,-1};
int mark[M][M];

int find(Pos start, Pos end);
/***********************************************************/
int main(void)
{
	int test_case;
	int T=0,th=0;
	/*
	   The freopen function below opens input.txt file in read only mode, and afterward,
	   the program will read from input.txt file instead of standard(keyboard) input.
	   To test your program, you may save input data in input.txt file,
	   and use freopen function to read from the file when using scanf function.
	   You may remove the comment symbols(//) in the below statement and use it.
	   But before submission, you must remove the freopen function or rewrite comment symbols(//).
	 */
	 freopen("input.txt", "r", stdin);
	/*
	   If you remove the statement below, your program's output may not be rocorded
	   when your program is terminated after the time limit.
	   For safety, please use setbuf(stdout, NULL); statement.
	 */
	setbuf(stdout, NULL);

	scanf("%d", &T);

	for(test_case = 0; test_case < T; test_case++)
	{
		int ret=0,i=0,j=0;
		/**********************************
		 * Implement your algorithm here. */
		scanf("%d", &th);
		
		for(i=0;i<M;i++)
		for(j=0;j<M;j++)
		{
			scanf("%d",&data[i][j]);
			mark[i][j]=0;
		}

		for(i=0;i<M*M;i++)
		{
			cross_store[i].x=-1;
			cross_store[i].y=-1;

			path_store[i].x=-1;
			path_store[i].y=-1;
		}
		cross_increase=0;
		path_increase=0;
		len=0;
		start.x=0;
		start.y=0;
		end.x=9;
		end.y=9;

		ret=find(start,end);
		/*************************************/
		// Print the answer to standard output(screen).
		printf("#%d %d\n",th,ret);
	    fflush(stdout);//修复Eclipse printf（）不能显示的小bug  
	    i=0;
	    while(path_store[i].x !=-1)
	    {
	    	printf("The %dth stp is (%d,%d) \n",i+1,path_store[i].x,path_store[i].y);
	    	fflush(stdout);//修复Eclipse printf（）不能显示的小bug
	    	i++;
	    }

	}
	return (0);//Your program should return 0 on normal termination.
}

int find(Pos start, Pos end)
{
	int x=0,y=0,tx=0,ty=0,k=0,flag=0;
	x=start.x;
	y=start.y;
	while(1)
	{
		path_store[path_increase].x=x;
		path_store[path_increase].y=y;
		path_store[path_increase].len=len++;
		path_increase++;

		mark[x][y]=1;

		if(x==end.x && y==end.y)//找到
			return (1);
		flag=0;
		for(k=0;k<4;k++)// check cross
		{
			tx=x+offset[k][0];
			ty=y+offset[k][1];
			if( tx>=0 && tx<M && ty>=0 && ty<M && mark[tx][ty]==0 )
			{
				if(data[tx][ty]==1)
					flag++;
			}
		}
		if(flag==0)//no path; return last cross
		{
			x=cross_store[cross_increase-1].x;
			y=cross_store[cross_increase-1].y;
			len=cross_store[cross_increase-1].len;
			path_increase=len;
		}
		if(flag>1)// encounter cross ,store
		{
			cross_store[cross_increase].x=x;
			cross_store[cross_increase].y=y;
			cross_store[cross_increase].len= len;
			cross_increase++;
		}
		for(k=0;k<4;k++)// go
		{
			tx=x+offset[k][0];
			ty=y+offset[k][1];
			if( tx>=0 && tx<M && ty>=0 && ty<M && mark[tx][ty]==0 )
			{
				if(data[tx][ty]==1)
				{
					x=tx;
					y=ty;
					//continue;
					break;
				}
			}
		}

	}
}