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 |
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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。 |
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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！ |
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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"。
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Sample Input
3 4 1 2 3 4 0 0 0 0 4 3 2 1 4 1 1 3 4 1 1 2 4 1 1 3 3 2 1 2 4 3 4 0 1 4 3 0 2 4 1 0 0 0 0 2 1 1 2 4 1 3 2 3 0 0 |
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Sample Output
YES NO NO NO NO YES |
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Code
1 #include <stdio.h> 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 = 1 11 */ 12 int map[N + 5][N + 5], n, m, ex, ey, flag; 13 14 /* 15 dx[5]: the x of the directory 16 dy[5]: the y of the directory 17 such as, (dx[1], dy[1]), that is (0, -1), means moving left.(maybe u will feel confuse, but it is) 18 dir: directory 19 dir = 1, means left 20 dir = 2, means up 21 dir = 3, means down 22 dir = 4, means right 23 bx: begin of x 24 by: begin of y 25 start point is (bx, by) 26 ct: count of turning 27 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 desitanation 35 if (flag == 1) return; 36 //mean u turn to many 37 if (ct >= 3) return; 38 //this is crucial, without this, u will wrong 39 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 index 43 if(bx <= 0 || bx > n || by <= 0 || by > m) 44 return; 45 //arrive the destination, note that dir =0 means this is the first time 46 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 interger 52 if (dir != 0 && map[bx][by] != 0) 53 return; 54 //from all of the directions 55 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 query 66 int q, i, j, bx, by; 67 while (scanf("%d %d", &n, &m) && (m || n)) 68 { 69 //input 70 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 different 78 if (bx == ex && bx == ey || map[bx][by] == 0 || map[bx][by] == 0 || map[bx][by] != map[ex][ey]) 79 puts("NO"); 80 else 81 { 82 flag = 0; 83 dfs (bx, by, 0, 0); 84 if (flag == 1) 85 puts("YES"); 86 else 87 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.
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