hud 4740 The Donkey of Gui Zhou（深搜DFS）

The Donkey of Gui Zhou
Time Limit:1000MS Memory Limit:32768KB
64bit IO Format:%I64d & %I64u

Description
There was no donkey in the province of Gui Zhou, China. A trouble maker shipped
one and put it in the forest which could be considered as an N×N grid. The coordinates
of the up-left cell is (0,0) , the down-right cell is(N-1,N-1) and he
cell below the up-left
cell is (1,0)..... A 4×4 grid is shown below:



The donkey lived happily until it saw a tiger far away. The donkey had never seen
a tiger ,and the tiger had never seen a donkey. Both of them were frightened and wanted
to escape from each other. So they started running fast. Because they were scared, they
were running in a way that didn't make any sense. Each step they moved to the next cell
in their running direction, but they couldn't get out of the forest. And because they both
wanted to go to new places, the donkey would never stepped into a cell which had already
been visited by itself, and the tiger acted the same way. Both the donkey and the tiger ran
in a random direction at the beginning and they always had the same speed. They would
not change their directions until they couldn't run straight ahead any more. If they couldn't
go ahead any more ,they changed their directions immediately. When changing direction,
the donkey always turned right and the tiger always turned left. If they made a turn and still
couldn't go ahead, they would stop running and stayed where they were, without trying to
make another turn. Now given their starting positions and directions, please count whether
they would meet in a cell.

Input

There are several test cases.

In each test case:
First line is an integer N, meaning that the forest is a N×N grid.
The second line contains three integers R, C and D, meaning that the donkey is in the
cell (R,C) when they started running, and it's original direction is D. D can be 0, 1, 2 or
3. 0 means east, 1 means south , 2 means west,and 3 means north.

The third line has the same format and meaning as the second line, but it is for the tiger.

The input ends with N = 0. ( 2 <= N <= 1000, 0 <= R, C < N)

Output
For each test case, if the donkey and the tiger would meet in a cell, print the coordinate
of the cell where they meet first time. If they would never meet, print -1 instead.

Sample Input

20 0 00 1 2040 1 3 2 00

Sample Output

-11 3

题意：有一个N*N的方格，a donkey和a tiger分别在一个小格子里面，donkey只能往前或
往右走，tiger只能往前或 往左走。同样地，当不能转弯时，就停下来。他们没走一格，
所花的时间一样。输出他们第一次相遇的地点，即格子的坐标。

代码：
#include <cstdio>

#include <cstring>

using namespace std;

#define MAXN 1005

int dir[4][2]= {{0,1},{1,0},{0,-1},{-1,0}};//东南西北

int n,p,q;

int donkey[MAXN][MAXN],tiger[MAXN][MAXN];

void DFS(int a,int b,int c,int x,int y,int z)

{

donkey[a][b]=1;

tiger[x][y]=1;

if(a==x&&b==y)

{

printf("%d %d\n",a,b);

return ;

}

if(p&&q)

{

printf("-1\n");

return ;

}

int aa,bb,xx,yy;

if(p)

{

aa=a;

bb=b;

}

else

{

aa=a+dir[c][0];

bb=b+dir[c][1];

if(aa<0||bb<0||aa>=n||bb>=n||donkey[aa][bb]==1)

{

c=(c+1)%4;//向右

aa=a+dir[c][0];

bb=b+dir[c][1];

if(aa<0||bb<0||aa>=n||bb>=n||donkey[aa][bb]==1)

{

p=1;

aa=a;

bb=b;

}

}

}

if(q)

{

xx=x;

yy=y;

}

else

{

xx=x+dir[z][0];

yy=y+dir[z][1];

if(xx<0||yy<0||xx>=n||yy>=n||tiger[xx][yy]==1)

{

z=(z-1+4)%4;//向左

xx=x+dir[z][0];

yy=y+dir[z][1];

if(xx<0||yy<0||xx>=n||yy>=n||tiger[xx][yy]==1)

{

q=1;

xx=x;

yy=y;

}

}

}

DFS(aa,bb,c,xx,yy,z);

}

int main()

{

int a,b,c,x,y,z;

while(scanf("%d",&n),n)

{

scanf("%d%d%d",&a,&b,&c);

scanf("%d%d%d",&x,&y,&z);

if(a==x&&b==y)

{

printf("%d %d\n",a,b);

continue;

}

else

{

memset(donkey,0,sizeof(donkey));

memset(tiger,0,sizeof(tiger));

p=q=0;

DFS(a,b,c,x,y,z);

}

}

return 0;

}