POJ 1204 Word Puzzles (AC自动机)
来源：程序员人生 发布时间：2015-04-11 09:46:24 阅读次数：2797次
Word Puzzles
| Time Limit: 5000MS |
|
Memory Limit: 65536K |
| Total Submissions: 9926 |
|
Accepted: 3711 |
|
Special Judge |
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
0 15 G
2 11 C
7 18 A
4 8 C
16 13 B
4 15 E
10 3 D
5 1 E
19 7 C
11 11 H
Source
Southwestern Europe 2002
题目链接：http://poj.org/problem?id=1204
题目大意：给1个字符矩阵，和1些单词，求这些单词的首字母的位置和单词的方向，1共8个方向，正北为A，顺时针BCD...
题目分析：对单词建立trie树，构造ac自动机，以4个边上的字符为出发点向8个方向枚举找
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int const MAX = 1005;
char map[MAX][MAX], s[MAX];
int r, c, n;
int len[MAX], ans[MAX][3];
int dx[8] = {⑴, ⑴, 0, 1, 1, 1, 0, ⑴};
int dy[8] = {0, 1, 1, 1, 0, ⑴, ⑴, ⑴};
struct node
{
int id;
node *next[26];
node *fail;
node()
{
id = ⑴;
memset(next, NULL, sizeof(next));
fail = NULL;
}
};
void Insert(node *p, char *s, int id)
{
for(int i = 0; s[i] != ' '; i++)
{
int idx = s[i] - 'A';
if(p -> next[idx] == NULL)
p -> next[idx] = new node();
p = p -> next[idx];
}
p -> id = id;
}
void AC_Automation(node *root)
{
queue <node*> q;
q.push(root);
while(!q.empty())
{
node *p = q.front();
q.pop();
for(int i = 0; i < 26; i++)
{
if(p -> next[i])
{
if(p == root)
p -> next[i] -> fail = root;
else
p -> next[i] -> fail = p -> fail -> next[i];
q.push(p -> next[i]);
}
else
{
if(p == root)
p -> next[i] = root;
else
p -> next[i] = p -> fail -> next[i];
}
}
}
}
bool judge(int i, int j)
{
if(i >= 0 && i < r && j >= 0 && j < c)
return true;
return false;
}
void Query(node *root, int x, int y, int k)
{
node *p = root;
for(int i = x, j = y; judge(i, j); i += dx[k], j += dy[k])
{
int idx = map[i][j] - 'A';
while(!p -> next[idx] && p != root)
p = p -> fail;
p = p -> next[idx];
if(!p)
{
p = root;
continue;
}
node *tmp = p;
while(tmp != root)
{
//找到单词，则记录下来相干信息
if(tmp -> id != ⑴)
{
int id = tmp -> id;
ans[id][0] = i - len[id] * dx[k];
ans[id][1] = j - len[id] * dy[k];
ans[id][2] = k;
tmp -> id = ⑴;
}
else
break;
tmp = tmp -> fail;
}
}
}
int main()
{
scanf("%d %d %d", &r, &c, &n);
node *root = new node();
for(int i = 0; i < r; i++)
scanf("%s", map[i]);
for(int i = 0; i < n; i++)
{
scanf("%s", s);
len[i] = strlen(s) - 1;
Insert(root, s, i);
}
AC_Automation(root);
//枚举各个边界的各个方向
for(int i = 0; i < c; i++)
{
for(int dir = 0; dir < 8; dir++)
{
Query(root, 0, i, dir);
Query(root, r - 1, i, dir);
}
}
for(int i = 0; i < r; i++)
{
for(int dir = 0; dir < 8; dir++)
{
Query(root, i, 0, dir);
Query(root, i, c - 1, dir);
}
}
for(int i = 0; i < n; i++)
printf("%d %d %c
", ans[i][0], ans[i][1], ans[i][2] + 'A');
}
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