poj1204之AC自动机

Word Puzzles

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8235		Accepted: 3104		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

题意:输入n,m,w代表先给定n行m列字符,接下来w行每行给定一组字符串,求该字符串在n*m的矩阵中首先出现的起始位置,字符串可以和矩阵中以某点开始8个方向进行匹配

输出首先匹配的起始点坐标和匹配的方向

方向为A,B,C,D...

分析:将w各字符串插入字典树,然后用n*m的矩阵去匹配,匹配方法是枚举8个方向去匹配

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=1000+10;
char s[MAX][MAX],b[MAX];
int n,m,w;
int dir[8][2]={0,1,0,-1,1,0,-1,0,1,1,-1,-1,1,-1,-1,1};//八个方向
char ch[9]="CGEADHFB";
int pos[MAX][3];

struct TrieNode{
	int id;//记录第几个字符串 
	TrieNode *next[26],*fail;
	TrieNode(){
		id=0;
		fail=0;
		memset(next,0,sizeof next);
	}
}*root;

void InsertNode(char *a,int id){
	int len=strlen(a)-1;
	TrieNode *p=root;
	while(len>=0){//这里将a数组倒着插入字典树,方便匹配时记录匹配的终点即原串的起始点
		if(!p->next[a[len]-'A'])p->next[a[len]-'A']=new TrieNode;
		p=p->next[a[len--]-'A'];
	}
	p->id=id;
}

void Build_AC(){
	TrieNode *p=root,*next;
	queue<TrieNode *>q;
	q.push(root);
	while(!q.empty()){
		p=q.front();
		q.pop();
		for(int i=0;i<26;++i){
			if(p->next[i]){
				next=p->fail;
				while(next && !next->next[i])next=next->fail;
				if(next)p->next[i]->fail=next->next[i];
				else p->next[i]->fail=root;
				q.push(p->next[i]);
			}
		}
	}
}

void SearchTrie(int x,int y,int d,int id){
	TrieNode *p=root,*next; 
	while(x>=0 && y>=0 && x<n && y<m){
		while(p && !p->next[s[x][y]-'A'])p=p->fail;
		if(!p)p=root;
		else p=p->next[s[x][y]-'A'];
		next=p;
		while(next != root){
			if(next->id){//记录原串被匹配的起始点
				int k=next->id;
				if(pos[k][0]>x || (pos[k][0] == x && pos[k][1]>y)){
					pos[k][0]=x,pos[k][1]=y,pos[k][2]=id;
				} 
			} 
			next=next->fail; 
		}
		x+=dir[d][0];
		y+=dir[d][1];
	}
}

void Free(TrieNode *p){
	for(int i=0;i<26;++i)if(p->next[i])Free(p->next[i]);
	delete p;
}

int main(){
	while(cin>>n>>m>>w){
		root=new TrieNode;
		for(int i=0;i<n;++i)cin>>s[i];
		for(int i=1;i<=w;++i){
			cin>>b;
			InsertNode(b,i);
			pos[i][0]=pos[i][1]=INF;
		}
		Build_AC();
		for(int i=0;i<n;++i){
			SearchTrie(i,0,0,1),SearchTrie(i,m-1,1,0);//匹配左右方向
			SearchTrie(i,0,7,6),SearchTrie(i,m-1,6,7);//匹配左上部分的右上角和右下部分左下角
			SearchTrie(i,0,4,5),SearchTrie(i,m-1,5,4);//匹配左下部分右下角和右上部分左上角
		}
		for(int i=0;i<m;++i){
			SearchTrie(0,i,2,3),SearchTrie(n-1,i,3,2);//匹配上下方向
			SearchTrie(0,i,6,7),SearchTrie(n-1,i,7,6);//匹配左上部分左下角和右下部分右上角
			SearchTrie(0,i,4,5),SearchTrie(n-1,i,5,4);//匹配右上部分的右下角和左下部分左上角 
		}
		for(int i=1;i<=w;++i)cout<<pos[i][0]<<' '<<pos[i][1]<<' '<<ch[pos[i][2]]<<endl;
		Free(root);
	}
	return 0;
}