poj 1204 Word Puzzles 给出一个N*L的字符矩阵，再给出M个字符串，问这M个字符串在这个字符矩阵中出现的位置 Tire树-优快云博客

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/6342632

本文介绍了一种程序设计方法，旨在通过计算机自动解决大型单词拼图谜题，提升娱乐性和效率。详细阐述了输入解析、字典树构建、搜索算法以及输出结果的生成过程。

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Description

Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using table covers with word puzzles printed on them, possibly with the intent to minimise their client's perception of any possible delay in bringing them their order.

Even though word puzzles may be entertaining to solve by hand, they may become boring when they get very large. Computers do not yet get bored in solving tasks, therefore we thought you could devise a program to speedup (hopefully!) solution finding in such puzzles.

The following figure illustrates the PizzaHut puzzle. The names of the pizzas to be found in the puzzle are: MARGARITA, ALEMA, BARBECUE, TROPICAL, SUPREMA, LOUISIANA, CHEESEHAM, EUROPA, HAVAIANA, CAMPONESA.
Your task is to produce a program that given the word puzzle and words to be found in the puzzle, determines, for each word, the position of the first letter and its orientation in the puzzle.

You can assume that the left upper corner of the puzzle is the origin, (0,0). Furthemore, the orientation of the word is marked clockwise starting with letter A for north (note: there are 8 possible directions in total).

Input

The first line of input consists of three positive numbers, the number of lines, 0 < L <= 1000, the number of columns, 0 < C <= 1000, and the number of words to be found, 0 < W <= 1000. The following L input lines, each one of size C characters, contain the word puzzle. Then at last the W words are input one per line.

Output

Your program should output, for each word (using the same order as the words were input) a triplet defining the coordinates, line and column, where the first letter of the word appears, followed by a letter indicating the orientation of the word according to the rules define above. Each value in the triplet must be separated by one space only.

Sample Input

20 20 10
QWSPILAATIRAGRAMYKEI
AGTRCLQAXLPOIJLFVBUQ
TQTKAZXVMRWALEMAPKCW
LIEACNKAZXKPOTPIZCEO
FGKLSTCBTROPICALBLBC
JEWHJEEWSMLPOEKORORA
LUPQWRNJOAAGJKMUSJAE
KRQEIOLOAOQPRTVILCBZ
QOPUCAJSPPOUTMTSLPSF
LPOUYTRFGMMLKIUISXSW
WAHCPOIYTGAKLMNAHBVA
EIAKHPLBGSMCLOGNGJML
LDTIKENVCSWQAZUAOEAL
HOPLPGEJKMNUTIIORMNC
LOIUFTGSQACAXMOPBEIO
QOASDHOPEPNBUYUYOBXB
IONIAELOJHSWASMOUTRK
HPOIYTJPLNAQWDRIBITG
LPOINUYMRTEMPTMLMNBO
PAFCOPLHAVAIANALBPFS
MARGARITA
ALEMA
BARBECUE
TROPICAL
SUPREMA
LOUISIANA
CHEESEHAM
EUROPA
HAVAIANA
CAMPONESA

Sample Output

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
//给出一个N*L的字符矩阵，再给出M个字符串，问这M个字符串在这个字
//符矩阵中出现的位置。
//逆向思维
//先将M个字符串构造成字典树，然后再字符矩阵上枚举

const int maxn=1100;
struct trie
{
int num;
int index;
trie* next[26];
};
//静态链表
trie temp[1000010];
int tp;
//.......
trie* root;//trie树根节点
int c,l;//主串行列
int n;//模式串个数
char mat[maxn][maxn];//主串
char str[maxn];//模式串
int move[8][2]={-1,0,-1,1,0,1,1,1,1,0,1,-1,0,-1,-1,-1};
struct node//结果
{
int x,y,d;//头字母位置方向
};
node ans[maxn];
void reset(trie* p)
{
p->num=0,p->index=-1;
for(int i=0;i<26;i++) p->next[i]=NULL;
}
void init()
{
tp=0;
root=&temp[tp++];
reset(root);
for(int i=0;i<maxn;i++) ans[i].d=-1;
}
void insert(char* word,int index)
{
trie* p=root;
int i=0;
while(word[i])
{
  int x=word[i]-'A';
  if(p->next[x]==NULL)
  {
   p->next[x]=&temp[tp++];
   reset(p->next[x]);
  }
  p=p->next[x];
  i++;
}
p->num++,p->index=index;
}
void check(int x,int y,int d)
{
int i=x,j=y;
trie* p=root;
while(i>=0&&i<c&&j>=0&&j<l&&p!=NULL)
{
  int id=mat[i][j]-'A';
  trie* cnt=p->next[id];
  if(cnt!=NULL&&cnt->num&&ans[cnt->index].d==-1)
  {
   ans[cnt->index].x=x;
   ans[cnt->index].y=y;
   ans[cnt->index].d=d;
  }
  i+=move[d][0],j+=move[d][1];
  p=cnt;
}
}
void solve()
{
for(int i=0;i<c;i++)
{
  for(int j=0;j<l;j++)
  {
   for(int k=0;k<8;k++)
   {
    check(i,j,k);
   }
  }
}
for(int i=1;i<=n;i++) printf("%d %d %c/n",ans[i].x,ans[i].y,ans[i].d+'A');
}
int main()
{
init();
scanf("%d%d%d",&c,&l,&n);
for(int i=0;i<c;i++) scanf("%s",mat[i]);
for(int i=1;i<=n;i++)
{
  scanf("%s",str);
  insert(str,i);
}
solve();
return 0;
}