POJ 1204(AC自动机)

Word Puzzles

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 6543 Accepted: 2486 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. 

这是我做的第二道AC自动机的题（第一道是HD2222）。。没想到能1Y。。舒坦。。。我是分方向分横纵枚举。。。

#include<stdio.h>#include<string.h>#define maxn 1000001#define maxn2 1002#define maxn3 100#define maxn4 2000000#define kind 26#define clear(a,b) memset(a,b,sizeof(a))const int di[8][2] = {{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};const char tranf[8] = {'E','F','G','H','A','B','C','D'};char ma[maxn2][maxn2];int X[maxn2],Y[maxn2],dir[maxn2];struct AC_node{  int a,b;  struct AC_node *next[kind],*fail;  AC_node(){    fail = NULL;    a = 0;    b = 0;    clear(next,NULL);    }  } *q[maxn];char s1[maxn3],txt[maxn4];void AC_trie(struct AC_node *T,char s[],int t){  int k,i = strlen(s)-1;  struct AC_node *op = T;  while(i >= 0) {    k = s[i] - 'A';    if (op->next[k] == NULL) op->next[k] = new AC_node();    op = op->next[k];    i--;    }    op->a++;    op->b = t;}void AC_build(struct AC_node *T){  int i,h = 0,t = 0;  struct AC_node *op,*tt;  clear(q,NULL);  T->fail = NULL;  q[t++] = T;  while (h < t) {    op = q[h++];    for(i = 0;i < kind;i++) {      if (op->next[i]!=NULL) {        if (op == T) op->next[i]->fail = T;          else {            tt = op->fail;            while(tt != NULL) {              if (tt->next[i] != NULL) {                op->next[i]->fail = tt->next[i];                break;                }              tt = tt->fail;              }            if (tt == NULL) op->next[i]->fail = T;            }        q[t++] = op->next[i];        }      }    }}int AC_query(struct AC_node *T,int x,int y,int d){  int i = 0,ans = 0,k;  struct AC_node *op = T;  while(ma[x][y]) {    k = ma[x][y] - 'A';    while(op->next[k] == NULL && op != T) op = op->fail;    op = op->next[k];    op = (op == NULL)?T:op;    struct AC_node *tmp = op;    while(tmp != T) {      if (tmp->a > 0) {        if (x < X[tmp->b] || (x == X[tmp->b] && y < Y[tmp->b])) {          X[tmp->b] = x;          Y[tmp->b] = y;          dir[tmp->b] = d;          }        }      tmp = tmp->fail;      }    x+=di[d][0];    y+=di[d][1];    }  return ans;}int main(){  int j,m,n,i,k;  scanf("%d%d%d",&m,&n,&k);  clear(ma,0);  getchar();  for(i = 1;i <= m;i++,getchar())    for(j = 1;j <= n;j++) scanf("%c",&ma[i][j]);  struct AC_node *root = new AC_node();  for(i = 1;i <= k;i++) {      X[i] = maxn;      Y[i] = maxn;      scanf("%s",s1);      AC_trie(root,s1,i);    }  AC_build(root);  for(i = 1;i <= m;i++) {AC_query(root,i,1,2);AC_query(root,i,n,6);}  for(i = 1;i <= n;i++) {AC_query(root,1,i,4);AC_query(root,m,i,0);}  for(i = 1;i <= n;i++) {AC_query(root,1,i,3);AC_query(root,1,i,5);AC_query(root,m,i,7);AC_query(root,m,i,1);}  for(i = 1;i <= m;i++) {AC_query(root,i,1,1);AC_query(root,i,1,3);AC_query(root,i,n,5);AC_query(root,i,n,7);}  for(i = 1;i <= k;i++) printf("%d %d %c\n",X[i]-1,Y[i]-1,tranf[dir[i]]);  return 0;}