经典c程序(0033) ---迷宫通路求解(单条通路DFS)

/************************************************************************************** 
* Function     : test
* Create Date  : 2014/07/11
* Author       : NTSK13 
* Email        : beijiwei@qq.com 
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* Version      : V0.1                    
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经典c程序(0033) ---迷宫通路求解01

题目：较复杂迷宫求解  10*10 有多个分叉口,但只有一条通路

1.从起点到终点, 依次打印出通路坐标
2.输出通路地图,可行处标记为2

0为墙,1为通道

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

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

Pos start,end;

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

int check(Pos start,Pos end);
Pos store_pos[M*M];
void get_pos(Pos head);
int tail=0;
int mark_pos[M][M];
/***********************************************************/
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++)
		{
			store_pos[i].x=-1;
			store_pos[i].y=-1;
		}
		start.x=0;
		start.y=0;
		end.x=9;
		end.y=9;
		tail=0;
		ret=check(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  

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

int check(Pos start,Pos end)
{
	int x=0,y=0,tx=0,ty=0,k=0,ret=0;
	int i=0,j=0;
	Pos tmp;
	x=start.x;
	y=start.y;
	data[x][y]=2;

	mark[x][y]=1;

	if(x==end.x && y==end.y)
	{
		tmp.x=0;
		tmp.y=0;
		for(i=0;i<M;i++)
		for(j=0;j<M;j++)
			mark_pos[i][j]=0;
		get_pos(tmp);
		i=0;
		while( store_pos[i].x !=-1 )
		{
			printf("(%d,%d) ",store_pos[i].x,store_pos[i].y);
			fflush(stdout);
			i++;
		}
		printf("\n\n");
		fflush(stdout);
		for(i=0;i<M;i++)
		{
			for(j=0;j<M;j++)
			{
				printf("%d ",data[i][j]);
				fflush(stdout);
			}
			printf("\n");
			fflush(stdout);
		}
		printf("\n");
		fflush(stdout);
		return 1;
	}

	for(k=0;k<4;k++)
	{
		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)
			{
				start.x=tx;
				start.y=ty;
				ret=check(start,end);
				if(ret==1)
					return ret;
			}
		}
	}
	return ret;
}

void get_pos(Pos head)
{
	int x=0,y=0,tx=0,ty=0,k=0;
	x=head.x;
	y=head.y;
	store_pos[tail++]=head;
	mark_pos[x][y]=1;
	for(k=0;k<4;k++)
	{
		tx=x+offset[k][0];
		ty=y+offset[k][1];

		if( tx>=0 && tx<M && ty>=0 && ty<M && mark_pos[tx][ty]==0)
		{
			if(data[tx][ty]==2)
			{
				head.x=tx;
				head.y=ty;
				get_pos(head);
			}
		}
	}
}