UVAOJ 10010 —— Where's Waldorf? 搜索(照着一个方向一直搜)
2014-12-03 10:28
267 查看
Where's Waldorf? |
of letters, (
),
and a list of words, find the location in the grid at which the word can be found. A word matches a straight, uninterrupted line of letters in the grid. A word can match the letters in the grid regardless of case (i.e. upper and lower case letters are to be
treated as the same). The matching can be done in any of the eight directions either horizontally, vertically or diagonally through the grid.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followedby a blank line, and there is also a blank line between two consecutive inputs.
The input begins with a pair of integers, m followed by n,
in decimal
notation on a single line. The next m lines contain nletters each; this is the grid of letters in which the words of the list must be found. The letters in the grid may be in upper or lower case. Following the grid of letters, another integer k appears
on a line by itself (
). The next k lines of input contain the list of words to search for, one word per line.
These words may contain upper and lower case letters only (no spaces, hyphens or other non-alphabetic characters).
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.For each word in the word list, a pair of integers representing the location of the corresponding word in the grid must be output. The integers must be separated by a single space. The first integer is the line in the grid where
the first letter of the given word can be found (1 represents the topmost line in the grid, and m represents the bottommost line). The second integer is the column in the grid where the first letter of the given word can be found (1 represents the
leftmost column in the grid, and n represents the rightmost column in the grid). If a word can be found more than once in the grid, then the location which is output should correspond to the uppermost occurence of the word (i.e. the occurence which
places the first letter of the word closest to the top of the grid). If two or more words are uppermost, the output should correspond to the leftmost of these occurences. All words can be found at least once in the grid.
Sample Input
1 8 11 abcDEFGhigg hEbkWalDork FtyAwaldORm FtsimrLqsrc byoArBeDeyv Klcbqwikomk strEBGadhrb yUiqlxcnBjf 4 Waldorf Bambi Betty Dagbert
Sample Output
2 5 2 3 1 2 7 8
Miguel Revilla
2000-08-22
找准一个方向一直搜,不改变方向
#include <stdio.h>
#include <string.h>
#include <ctype.h>
char mp[55][55];
char cz[25];
int n,m;
int bhx[] = {-1,-1,0,1,1,1,0,-1};
int bhy[] = {0,1,1,1,0,-1,-1,-1};
int flag;
bool dfs(int x,int y,int i,int xb)
{
if(cz[xb + 1] == '\0')
{
flag = 1;
return true;
}
int nx = x + bhx[i];
int ny = y + bhy[i];
if(nx >= 0 && nx < n && ny >= 0 && ny < m && mp[nx][ny] == cz[xb + 1])
{
if(dfs(nx,ny,i,xb + 1))
return true;
}
return false;
}
bool dy(int x,int y)
{
int nx,ny,xb = 1;
for(int i = 0;i < 8;i++)
{
nx = x + bhx[i];
ny = y + bhy[i];
if(nx >= 0 && nx < n && ny >= 0 && ny < m && mp[nx][ny] == cz[1])
{
flag = 0;
dfs(nx,ny,i,xb);
if(flag)
{
printf("%d %d\n",x + 1,y + 1);
return true;
}
}
}
return false;
}
int main()
{
int t,k;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i = 0;i < n;i++)
scanf("%s",mp[i]);
for(int i = 0;i < n;i++)
{
for(int j = 0;j < m;j++)
mp[i][j] = tolower(mp[i][j]);
}
scanf("%d",&k);
while(k--)
{
scanf("%s",cz);
int len = strlen(cz);
for(int i = 0;i < len;i++)
cz[i] = tolower(cz[i]);
for(int i = 0;i < n;i++)
{
int j;
for(j = 0;j < m;j++)
{
if(mp[i][j] == cz[0])
{
if(len == 1)
{
printf("%d %d\n",i + 1,j + 1);
break;
}
else if(dy(i,j))
break;
}
}
if(j < m)
break;
}
}
if(t != 0)
printf("\n");
}
return 0;
}
相关文章推荐
- 小明同学喜欢体育锻炼,他常常去操场上跑步。跑道是一个圆形,在本题中,我们认为跑道是一个半径为R的圆形,设圆心的坐标原点(0,0)。小明跑步的起点坐标为(R,0),他沿着圆形跑道跑步,而且一直沿着一个方向跑步。回到家后,他查看了自己的计步器,计步器显示他跑步的总路程为L。小明想知道自己结束跑步时的坐标,但是他忘记自己是沿着顺时针方向还是逆时针方向跑的了。他想知道在这两种情况下的答案分别是多少。
- UvaOJ 10010 Where's Waldorf?
- UVA10010使用数组和一个for循环控制前进路径方向不变
- AspNetCommerce中的一个带分页和排序的搜索功能的存储过程
- 一个在文本文件搜索指定字符串的程序
- java中的基于套结字(socket)的通信 一. 一个双人单方向通信例子
- 一个有趣的查找--搜索最大值所在的ID号 (轉自:http://blog.csdn.net/dhlhh)
- 一个搜索脚本,想做搜索引擎的参考一下吧
- GIS图形的地理方向——一个常被忽视的问题
- ASP编写完整的一个IP所在地搜索类
- 一个树的递归搜索
- 搜索和替换文件或目录的一个好类--很实用
- 一直都不清楚RSS,照着一个C#的例子,随便写了一个,也不知道对不对!
- ASP编写完整的一个IP所在地搜索类
- 一个查找文件的类:它的长处在于能够搜索子目录并且是可控制的
- 一个方向控制射击小游戏的代码分析!(AS1.0)
- 一个简单的错误,一直没有仔细考虑。
- 写给公司的一个Bug需求管理系统,公司一直使用良好
- AspNetCommerce中的一个带分页和排序的搜索功能的存储过程
- 一个硬盘文件搜索的Asp源码