经典c程序(0034) ---迷宫通路求解(多条通路DFS)

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* 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(0034) ---迷宫通路求解02

题目：较复杂迷宫求解  10*10 有多个分叉口,多条通路

1.求通路个数
2.分别输出通路地图,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 0 0 1
1 1 1 0 1 0 0 1 1 1
0 0 1 1 1 1 1 1 0 1
0 0 1 0 1 1 0 1 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];

void check(Pos start,Pos end);
Pos store_pos[M*M];
void get_pos(Pos head);
int tail=0;
int mark_pos[M][M];

int count=0;
int flag=0;
/***********************************************************/
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 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;
		}

		start.x=0;
		start.y=0;
		end.x=9;
		end.y=9;

		count=0;
		check(start,end);
		/**********************************/
		// Print the answer to standard output(screen).
		printf("#%d %d\n",th,count);//个数同路
	    fflush(stdout);//修复Eclipse printf（）不能显示的小bug  

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

void check(Pos start,Pos end)
{
	int x=0,y=0,tx=0,ty=0,k=0;
	int i=0,j=0,value=0;
	Pos tmp;
	x=start.x;
	y=start.y;
	value=data[x][y];//保存当前位置值

	data[x][y]=2;
	mark[x][y]=1;

	if(x==end.x && y==end.y)
	{
		count++;
		tmp.x=0;
		tmp.y=0;
		for(i=0;i<M;i++)
		for(j=0;j<M;j++)
			mark_pos[i][j]=0;
		tail=0;
		for(i=0;i<M*M;i++)
		{
			store_pos[i].x=-1;
			store_pos[i].y=-1;
		}
		flag=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);
	/****************************************************/
		data[x][y]=value; //到达终点后 回溯
		mark[x][y]=0;
		return;
	/****************************************************/
	}

	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;
				check(start,end);
			}
		}
	}
	data[x][y]=value;
	mark[x][y]=0;
}

void get_pos(Pos head)
{
	int x=0,y=0,tx=0,ty=0,k=0;
	x=head.x;
	y=head.y;
	if(flag==1)//这个必须加,否则 输出(9,9) 之后,有可能输出其他点 (  9,5  9,4)
		return;
	store_pos[tail++]=head;
	mark_pos[x][y]=1;
	if(x==9 && y==9)
	{
		flag=1;
		return;
	}
	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);
			}
		}
	}
}