POJ 题目1204 Word Puzzles（AC自动机，多个方向查询）

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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

Output

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

题目大意：给你一个矩阵，然后问每个串在矩阵中的起点和方向，方向从上（A）顺时针8个方向

就是也8个方向的查询就好。。

ac代码

#include<stdio.h>
#include<string.h>
char map[1010][1010],key[1010];
int dx[8]={-1,-1,0,1,1,1,0,-1};
int dy[8]={0,1,1,1,0,-1,-1,-1};
int ansx[1010],ansy[1010],ansd[1010],len[1010];
int head,tail;
int n,m,w;
struct node  
{  
    node *fail;  
    node *next[26];  
    int id;
    node()  
    {  
        fail=NULL;  
        id=0;  
        for(int i=0;i<26;i++)  
            next[i]=NULL;  
    }  
}*q[50005000];  
node *root;  
void insert(char *s,int id)  
{  
    int temp,len,i;  
    node *p=root;  
    len=strlen(s);  
    for(i=0;i<len;i++)  
    {  
        temp=s[i]-'A';  
        if(p->next[temp]==NULL)  
            p->next[temp]=new node();  
        p=p->next[temp];  
    }  
    p->id=id;      
}  
void build_ac()  
{  
	head=tail=0;
    q[tail++]=root;  
    while(head!=tail)  
    {  
        node *p=q[head++];  
        node *temp=NULL;  
        for(int i=0;i<26;i++)  
        {  
            if(p->next[i]!=NULL)  
            {  
                if(p==root)  
                    p->next[i]->fail=root;  
                else  
                {  
                    temp=p->fail;  
                    while(temp!=NULL)  
                    {  
                        if(temp->next[i]!=NULL)  
                        {  
                            p->next[i]->fail=temp->next[i];  
                            break;  
                        }  
                        temp=temp->fail;  
                    }  
                    if(temp==NULL)  
                    {  
                        p->next[i]->fail=root;  
                    }  
                }  
                q[tail++]=p->next[i];  
            }  
        }  
    }  
}
void query(int x,int y,int dir)
{
	node *p=root,*temp;
	int i,j;
	for(i=x,j=y;i>=0&&i<n&&j>=0&&j<m;i+=dx[dir],j+=dy[dir])
	{
		int x=map[i][j]-'A'; 
        while(p->next[x]==NULL&&p!=root)  
            p=p->fail;  
        p=p->next[x];  
        if(p==NULL)  
        {  
            p=root;  
        }  
        temp=p;  
        while(temp!=root&&temp->id!=-1)  
        {  
            int id=temp->id;
			ansx[id]=i-(len[id]-1)*dx[dir];
			ansy[id]=j-(len[id]-1)*dy[dir];
			ansd[id]=dir;
			temp->id=-1;
            temp=temp->fail;  
        }  
	}
}
int main()
{
//	int n,m,w;
	while(scanf("%d%d%d",&n,&m,&w)!=EOF)
	{
		int i,j;
		root=new node();
		for(i=0;i<n;i++)
		{
			scanf("%s",map[i]);
		}
		for(i=1;i<=w;i++)
		{
			scanf("%s",key);
			len[i]=strlen(key);
			insert(key,i);
		}
		build_ac();
		for(i=0;i<m;i++)
		{
			for(j=0;j<8;j++)
			{
				query(0,i,j);
				query(n-1,i,j);
			}
		}
		for(i=0;i<n;i++)
		{
			for(j=0;j<8;j++)
			{
				query(i,0,j);
				query(i,m-1,j);
			}
		}
		for(i=1;i<=w;i++)
		{
			printf("%d %d %c\n",ansx[i],ansy[i],ansd[i]+'A');
		}
	}
}

POJ 题目1204 Word Puzzles（AC自动机，多个方向查询）

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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/47439299