Word Puzzles （AC自动机）

最新推荐文章于 2021-04-22 22:23:31 发布

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最新推荐文章于 2021-04-22 22:23:31 发布

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分类专栏： ac自动机

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本文介绍了一种使用AC自动机算法快速解决大型单词搜索谜题的方法，通过构建AC自动机加速在字符矩阵中查找指定单词的过程，适用于各种规模的谜题，尤其在处理大量单词时效率显著。

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https://cn.vjudge.net/problem/POJ-1204

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

题目大意：给一个L*C字符矩阵和W个字符串，问那些字符串出现在矩阵的位置，横竖斜八个向。

枚举字符矩阵八个方向的所有字符串构成主串，然后在W个模式串构造的AC自动机上跑。

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn=2e6+10;
const int maxc=26;
char mp[1005][1005];
char s[1005];
int ans[1005][3];
int Len[1005];
bool vis[1005];
int dx[8]={0,1,0,-1,1,1,-1,-1};
int dy[8]={1,0,-1,0,-1,1,1,-1};
char dir[8]={'C','E','G','A','F','D','B','H'};
int l,c,w;
int root=1;
int tot=0;
struct TRIE{
	int next[maxn][maxc],fail[maxn],num[maxn];
	
	char base;
	void init(){
		for(int i=1;i<=tot;i++){
			memset(next[i],0,sizeof(next[i]));
			fail[i]=0;
			num[i]=0;
		}
		tot=1;
	}
	void add(char s[],int id){
		int len=strlen(s);
		int now=root;
		for(int i=0;i<len;i++){
			int c=s[i]-base;
			if(!next[now][c]) next[now][c]=++tot;
			now=next[now][c];
		}
		num[now]=id;
	}
	void getfail(){
		queue<int> q; 
		fail[root]=root;
		for(int i=0;i<26;i++){
			if(!next[root][i]) next[root][i]=root;
			else{
				fail[next[root][i]]=root;
				q.push(next[root][i]);
			}	
		}
		while(!q.empty()){
			int now=q.front();
			q.pop(); 
			for(int i=0;i<26;i++){
				if(!next[now][i]) 
				    next[now][i]=next[fail[now]][i];
				else{
					fail[next[now][i]]=next[fail[now]][i];
					q.push(next[now][i]);
				}
			}
			
		}	
	} 
	void check(int x,int y,int d){
		int now=root;
		while(x>=0&&x<l&&y>=0&&y<c){
			int c=mp[x][y]-base;
			now=next[now][c];
			int p=now;
			while(p!=root){
				if(num[p]&&!vis[num[p]]){
					vis[num[p]]=true;
					ans[num[p]][0]=x-dx[d]*(Len[num[p]]-1);
					ans[num[p]][1]=y-dy[d]*(Len[num[p]]-1);
					ans[num[p]][2]=d;
				}
				p=fail[p];
			} 
			x+=dx[d];
			y+=dy[d];
		}
	} 
	/*
	void find(char s[]){//求每种模式串在主串中的数目 
	    int len=strlen(s);
		int now=root;
		for(int i=0;i<len;i++){
			int c=s[i]-base;
			now=next[now][c];
			int p=now;
			while(p!=root){
				if(num[p]) ans[num[p]]++;
				p=fail[p];
			}
		} 
	}
	*/
	/* 
	void find(char s[]){//求主串中有哪几种模式串 
		cnt=0;
		int now=root;
		int len=strlen(s);
		for(int i=0;i<len;i++){
			int c=s[i]-base;
			now=next[now][c];
			int p=now;
			while(p!=root){
				if(num[p]) ans[cnt++]=num[p];
				p=fail[p];
			}
		}
	}
	*/
}tr;
int main(){
	scanf("%d%d%d",&l,&c,&w);
	for(int i=0;i<l;i++) scanf("%s",mp[i]);
	tr.init();
	tr.base='A';
	for(int i=1;i<=w;i++){
		scanf("%s",s);
		Len[i]=strlen(s);
		tr.add(s,i);
	}
	tr.getfail();
	for(int i=0;i<l;i++){
		for(int j=0;j<8;j++){
			tr.check(i,0,j);
		}
	}
	for(int i=0;i<l;i++){
		for(int j=0;j<8;j++){
			tr.check(i,c-1,j);
		}
	}
	for(int i=0;i<c;i++){
		for(int j=0;j<8;j++){
			tr.check(0,i,j);
		}
	}
	for(int i=0;i<c;i++){
		for(int j=0;j<8;j++){
			tr.check(l-1,i,j);
		}
	}
	for(int i=1;i<=w;i++)
        printf("%d %d %c\n",ans[i][0],ans[i][1],dir[ans[i][2]]);
	return 0;
}