连连看

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 126 Accepted Submission(s): 63

Problem Description

“连连看”相信很多人都玩过。没玩过也没关系，下面我给大家介绍一下游戏规则：在一个棋盘中，放了很多的棋子。如果某两个相同的棋子，可以通过一条线连起来（这条线不能经过其它棋子），而且线的转折次数不超过两次，那么这两个棋子就可以在棋盘上消去。不好意思，由于我以前没有玩过连连看，咨询了同学的意见，连线不能从外面绕过去的，但事实上这是错的。现在已经酿成大祸，就只能将错就错了，连线不能从外围绕过。

玩家鼠标先后点击两块棋子，试图将他们消去，然后游戏的后台判断这两个方格能不能消去。现在你的任务就是写这个后台程序。

 

Input

输入数据有多组。每组数据的第一行有两个正整数n,m(0<n<=1000,0<m<1000)，分别表示棋盘的行数与列数。在接下来的n行中，每行有m个非负整数描述棋盘的方格分布。0表示这个位置没有棋子，正整数表示棋子的类型。接下来的一行是一个正整数q(0<q<50)，表示下面有q次询问。在接下来的q行里，每行有四个正整数x1,y1,x2,y2,表示询问第x1行y1列的棋子与第x2行y2列的棋子能不能消去。n=0,m=0时，输入结束。

注意：询问之间无先后关系，都是针对当前状态的！

 

Output

每一组输入数据对应一行输出。如果能消去则输出"YES",不能则输出"NO"。

 

Sample Input

3 41 2 3 40 0 0 04 3 2 141 1 3 41 1 2 41 1 3 32 1 2 43 40 1 4 30 2 4 10 0 0 021 1 2 41 3 2 30 0

 

Sample Output

YESNONONONOYES

 

Code

 

 

1 #include 
          
            2 #define N 1000 3 /* 4     map[N + 5][N + 5]: the input matrix 5     n: the number of rows 6     m: the number of column 7     ex: end of x 8     ey: end of y 9     destination point is (ex, ey)10     flag: when can arrive the destination, flag = 111 */12 int map[N + 5][N + 5], n, m, ex, ey, flag;13 14 /*15     dx[5]: the x of the directory16     dy[5]: the y of the directory17     such as, (dx[1], dy[1]), that is (0, -1), means moving left.(maybe u will feel confuse, but it is)18     dir: directory19         dir = 1, means left20         dir = 2, means up21         dir = 3, means down22         dir = 4, means right23     bx: begin of x24     by: begin of y25     start point is (bx, by)26     ct: count of turning27     28 */29 int dx[5] = {
        0, 0, -1, 1, 0};30 int dy[5] = {
        0, -1, 0, 0, 1};31 void dfs(int bx, int by, int ct, int dir)32 {33     int i;34     //arrive the desitanation35     if (flag == 1)    return;36     //mean u turn to many 37     if (ct >= 3)    return;38     //this is crucial, without this, u will wrong39     if (ct == 2)40          if (dir == 1 && (bx != ex || by < ey) || dir == 2 && (bx < ex || by != ey) || dir == 3 && (bx > ex || by != ey) || dir == 4 && (bx != ex || by > ey))41             return;42     //out of index43     if(bx <= 0 || bx > n || by <= 0 || by > m)44         return;45     //arrive the destination, note that dir =0 means this is the first time46     if (dir != 0 && bx == ex && by == ey)47     {48         flag = 1;49         return;50     }51     //u will be blocked when u come across a non-zero interger52     if (dir != 0 && map[bx][by] != 0)53         return;54     //from all of the directions55     for (i = 1; i <= 4; i++)56     {57         if (dir == 0 || i == dir)58             dfs(bx + dx[i], by + dy[i], ct, i);59         else if (i + dir != 5)60             dfs(bx + dx[i], by + dy[i], ct + 1, i);61     }62 }63 void main()64 {    65     //q: the num of the query66     int q, i, j, bx, by;67     while (scanf("%d %d", &n, &m) && (m || n))68     {69         //input70         for (i = 1; i <= n; i++)71             for (j = 1; j <= m; j++)72                 scanf("%d", &map[i][j]);73             scanf("%d", &q);74         for (j = 1; j <= q; j++)75         {76             scanf("%d %d %d %d", &bx, &by, &ex, &ey);77             //the input points is same or there is a zero or the two numbers are different78             if (bx == ex && bx == ey || map[bx][by] == 0 || map[bx][by] == 0 || map[bx][by] != map[ex][ey])79                 puts("NO");80             else81             {82                 flag = 0;83                 dfs (bx, by, 0, 0);84                 if (flag == 1)85                     puts("YES");86                 else87                     puts("NO");88             }89         }90     }91 }
          

 

The code is in detail, what I want stress is that the row and column are different to the coodinate.