POJ1204-Word Puzzles
Word Puzzles
Time Limit: 5000MS
Memory Limit: 65536K
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
題意是在字母表中找W個單詞的位置，以正北爲A，順時針輸出方向。
觀察問題，發現是一個多串匹配的問題，即可用AC自動機求解。
AC自動機的程序過程：
（1）用模式串建立Trie樹，用數組實現浪費空間，不過時間上更快一點。
（2）在Trie樹構建類似KMP算法的pre後綴數組，又叫匹配失敗指針
這一步可以通過隊列實現。
這裏注意要將將pre指向結點的標記繼承，是一個對Trie樹的加強，以保證將主串中所有的模式串找到。
（3）在主串上順序遍歷，同時在AC自動機漫遊，就可以求出模式串在主串位置。
這道題只需要將四周的格子爲起點，八個方向跑一遍AC自動機即可。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 1000 + 5;
const int d1[10] = {0, -1, -1, 0, 1, 1, 1, 0, -1};
const int d2[10] = {0, 0, 1, 1, 1, 0, -1, -1, -1};
typedef vector<pair<int, int> > vec;
#define fs first
#define se second
struct Trie{
int tree[100000][30], pre[100000], tot;
bool flag[100000];
queue<int> q;
vector<int> id[100000];
void init() {
memset(tree, 0, sizeof tree);
memset(flag, 0, sizeof flag);
tot = 1;
}
void insert(char *s, int id0) {
int cur = 1;
for (int i = 0; s[i]; ++i) {
int tep = s[i] - 'A' + 1;
if (tree[cur][tep])
cur = tree[cur][tep];
else {
tree[cur][tep] = ++tot;
cur = tot;
}
}
flag[cur] = 1;
id[cur].push_back(id0);
}
void buildPre() {
q.push(1);
while (!q.empty()) {
int cur = q.front();
q.pop();
for (int i = 1; i <= 26; ++i) {
int tep = tree[cur][i];
if (tep) {
int j = pre[cur];
while (j && !tree[j][i]) j = pre[j];
if (!j) pre[tep] = 1;
else pre[tep] = tree[j][i];
if (flag[pre[tep]]) {
flag[tep] = 1;
for (int i = 0; i < id[pre[tep]].size(); ++i)
id[tep].push_back(id[pre[tep]][i]);
}
q.push(tep);
}
}
}
}
vec AC_automation(char *s) {
int cur = 1;
vec res;
for (int i = 0; s[i]; ++i) {
int tep = s[i] - 'A' + 1;
while (cur && !tree[cur][tep]) cur = pre[cur];
if (!cur) cur = 1;
if (tree[cur][tep]) cur = tree[cur][tep];
if (flag[cur]) res.push_back(make_pair(cur, i));
}
return res;
}
}AC;
char s[MAXN][MAXN], p[MAXN], bri[MAXN];
int len[MAXN], ans[MAXN][5], L, C, W;
void solve(int x, int y) {
for (int i = 1; i <= 8; ++i) {
memset(bri, 0, sizeof bri);
int dx = x, dy = y, tot = -1;
while (true) {
bri[++tot] = s[dx][dy];
dx += d1[i];
dy += d2[i];
if (dx<1 || dx>L || dy<1 || dy>C) break;
}
vec res = AC.AC_automation(bri);
for (int j = 0; j < res.size(); ++j)
for (int k = 0; k < AC.id[res[j].fs].size(); ++k) {
int cur = AC.id[res[j].fs][k];
int ps = res[j].se - len[cur] + 1;
ans[cur][0] = x + d1[i] * ps - 1;
ans[cur][1] = y + d2[i] * ps - 1;
ans[cur][2] = 'A' + i - 1;
}
}
}
int main()
{
freopen("in.txt", "r", stdin);
while(scanf("%d%d%d", &L, &C, &W) != EOF) {
AC.init();
for (int i = 1; i <= L; ++i)
scanf("%s", &s[i][1]);
for (int i = 1; i <= W; ++i) {
scanf("%s", p);
AC.insert(p, i);
len[i] = strlen(p);
}
AC.buildPre();
for (int i = 1; i <= L; ++i) {
solve(i, 1);
solve(i, C);
}
for (int i = 1; i <= C; ++i) {
solve(1, i);
solve(L, i);
}
for (int i = 1; i <= W; ++i)
printf("%d %d %c\n", ans[i][0], ans[i][1], (char)ans[i][2]);
}
return 0;
}