hdu4740 search (will explode stack, need to open manually)

http://acm.hdu.edu.cn/showproblem.php?pid=4740

Problem Descriptionthere is no donkey in the province of Gui Zhou, China. A trouble maker shipped one and put it in the forest which could is considered as an nxn grid. The coordinates of the Up-left cell is (0,0), the Down-right cell was (n-1,n-1) and the cell below the Up-left cell is (1, 0) ..... A 4x4 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 of the other. So they started running fast. Because They were scared, they were running in a-on-the-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 visite D by itself, and the tiger acted the same. Both the donkey and the Tiger ran in a random direction at the beginning and they all 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. 
Inputthere is several test cases.

In each test case:
First line was an integer N, meaning the forest is a nxn grid.

The second line contains three integers R, C and D, meaning the donkey are in the cell (r,c) when they started running , and it's original direction are 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 had the same format and meaning as the second line, but it was for the tiger.

The input ends with N = 0. (2 <= N <=, 0 <= R, C < N)
Outputfor Each test case, if the donkey and the Tiger would meet in a cell, print the coordinate of the cell where they me ET first time. If they would never meet, print-1 instead.
Sample Input

20 0 00 1 240 1 03 2 00

Sample Output

-11 3

Source2013 ACM/ICPC Asia Regional Hangzhou Online

/**hdu4740 search (will explode stack, need to open up manually) The topic: There is a donkey and a tiger, both in a n*n chess board run, to determine whether there is a point in time and the two ran to the same lattice. As for them to run to the way, or to read the question of the idea: two times Dfs with two vectors to save the donkey and the tiger in the position I time, find Vector1 I and Vector2 i inside coordinates first coincident on the line */#pragma comment (linker, "/ stack:10240000000000,10240000000000 ") #include <stdio.h> #include <string.h> #include <algorithm># Include <iostream> #include <vector>using namespace Std;vector <pair<int,int> > Vec,vec1;bool    Flag[1005][1005];int n;void dfs (int x,int y,int r,int num)//donkey left {flag[x][y]=1;    Vec.push_back (Make_pair (x, y));        if (r==1) {if (flag[x+1][y]==0&&x+1<=n) {DFS (x+1,y,r,num++);            } else {if (flag[x][y-1]==0&&y-1>0) {DFS (x,y-1,2,num++); }}} and else if (r==2) {if (flag[x][y-1]==0&&y-1>0) {DFS (x,y-1        , r,num++);            } else {if (flag[x-1][y]==0&&x-1>0) {    DFS (x-1,y,3,num++);         }}} and else if (r==3) {if (flag[x-1][y]==0&&x-1>0) {DFS (x-1,y,r,num++); } else {if (flag[x][y+1]==0&&y+1<=n) {DFS (x,y+1,0,n            um++);        }}}} else {if (flag[x][y+1]==0&&y+1<=n) {DFS (x,y+1,r,num++);            } else {if (flag[x+1][y]==0&&x+1<=n) {DFS (x+1,y,1,num++);    }}}}void dfs1 (int x,int y,int r,int num)///Tiger right {flag[x][y]=1;    Vec1.push_back (Make_pair (x, y));        if (r==1) {if (flag[x+1][y]==0&&x+1<=n) {DFS1 (x+1,y,r,num++);            } else {if (flag[x][y+1]==0&&y+1<=n) {DFS1 (x,y+1,0,num++);  }}} and else if (r==2) {if (flag[x][y-1]==0&&y-1>0) {          DFS1 (x,y-1,r,num++);            } else {if (flag[x+1][y]==0&&x+1<=n) {DFS1 (x+1,y,1,num++); }}} and else if (r==3) {if (flag[x-1][y]==0&&x-1>0) {DFS1 (x-1        , y,r,num++);            } else {if (flag[x][y-1]==0&&y-1>0) {DFS1 (x,y-1,2,num++); }}} and else {if (flag[x][y+1]==0&&y+1<=n) {DFS1 (x,y+1,r,num+        +);            } else {if (flag[x-1][y]==0&&x-1>0) {DFS1 (x-1,y,3,num++);        }}}}int Main () {while (~SCANF ("%d", &n)) {if (n==0) break;        int x,y,r;        scanf ("%d%d%d", &x,&y,&r);        memset (flag,0,sizeof (flag));        Vec.clear ();        DFS (x+1,y+1,r,0);        int m=vec.size (); /*for (int i=0;i<m;i++) {PrinTF ("%d%d\n", vec[i].first,vec[i].second);        }*/scanf ("%d%d%d", &x,&y,&r);        memset (flag,0,sizeof (flag));        Vec1.clear ();        DFS1 (x+1,y+1,r,0);        int m1=vec1.size ();        /*for (int i=0;i<m1;i++) {printf ("%d%d\n", vec1[i].first,vec1[i].second);        }*/bool cnt=0; for (int i=0; i<m&&i<m1&&cnt==0; i++) {if (Vec[i].first==vec1[i].first&&vec                [I].second==vec1[i].second) {x=vec[i].first-1;                Y=vec[i].second-1;            Cnt=1;                }} if (cnt==0) {if (M&LT;M1) {for (int i=m; i<m1; i++)                    {if (Vec[m-1].first==vec1[i].first&&vec[m-1].second==vec1[i].second)                        {x=vec1[i].first-1;                        Y=vec1[i].second-1;                    Cnt=1;       }         }} else {for (int i=m1; i<m; i++) {                        if (Vec[i].first==vec1[m1-1].first&&vec[i].second==vec1[m1-1].second) {                        x=vec[i].first-1;                        Y=vec[i].second-1;                    Cnt=1;        }}}} if (cnt==0) printf (" -1\n");    else printf ("%d%d\n", x, y); } return 0;}

hdu4740 search (will explode stack, need to open manually)