poj 1204 AC自动机

转于：http://hi.baidu.com/fujj314/blog/item/378425abe3a30df31f17a23d.html

题目大意：是给出一个单词表，然后给出几组字符串，要你求出字符串在单词表匹配的起始位置，和搜寻方向

解题思路：AC自动机，tries图，看到这么多个方向，多个单词，想到了多模式字符串匹配。

AC自动机，tries图的原理和建立参考这个文档，讲得很详细http://wenku.baidu.com/view/af921feb6294dd88d0d26b47.html

应该把字符串建立tries图，用单词表建立的话，每个位置每个方向的单词都得建立tires图~~不可取

建好tries图之后，枚举单词表每个起点，每个方向与tries图是否匹配

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int kind = 26;const int maxn = 1001;struct node{node *fail;int num;node *next[kind];node(){fail = NULL;num = -1;memset(next, 0, sizeof(next));}};int dir[8][2] = {{-1, 0,}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1}, {-1, -1}};char words[maxn][maxn], database[maxn][maxn];int l, c, w, id, len[maxn], ans[maxn][3];void insert(char *str, node *root, int n);void build(node *root);void search(node *root, int x, int y, int k);inline bool ok(int x, int y);int main(){    node *root = new node();id = 0;    scanf("%d %d %d", &l, &c, &w);    for(int i = 0; i < l; i++)        scanf("%s", database[i]);    for(int i = 0; i < w; i++)    {        scanf("%s", words[i]);        insert(words[i], root, i);    }    build(root);        for(int i = 0; i < l; i++)    {        for(int k = 0; k < 8; k++)        {            search(root, i, 0, k);            search(root, i, c - 1, k);        }    }        for(int j = 0; j < c; j++)    {        for(int k = 0; k < 8; k++)        {            search(root, 0, j, k);            search(root, l - 1, j, k);        }    }        for(int i = 0; i < w; i++)        printf("%d %d %c\n", ans[i][0], ans[i][1], 'A' + ans[i][2]);    return 0;}inline bool ok(int x, int y){    return (x >= 0 && x < l && y >= 0 && y < c);}void insert(char *str, node *root, int n){    int i, index;    len[n] = strlen(str);    node *p = root;    for(i = 0; i < len[n]; i++)    {        index = str[i] - 'A';        if(!p->next[index])            break;        p = p->next[index];    }        for(; i < len[n]; i++)    {        index = str[i] - 'A';        node *tmp = new node();        p->next[index] = tmp;        p = tmp;    }    p->num = id++;}void build(node *root){    queue<node *> que;    node *tmp;    root->fail = NULL;    que.push(root);    while(!que.empty())    {        node *p = que.front();        que.pop();        tmp = NULL;        for(int i = 0; i < kind; i++)        {            if(p->next[i] != NULL)            {                if(p == root)                    p->next[i]->fail = root;                else                {                    tmp = p->fail;                    while(tmp != NULL)                    {                        if(tmp->next[i] != NULL)                        {                            p->next[i]->fail = tmp->next[i];                            break;                        }                        tmp = tmp->fail;                    }                    if(!tmp)                        p->next[i]->fail = root;                }                que.push(p->next[i]);            }        }    }}void search(node *root, int x, int y, int k){    node *p = root, *q;    int index;    while(ok(x, y))    {        index = database[x][y] - 'A';        while(p->next[index] == NULL && p != root)            p = p->fail;        p = p->next[index];        if(p == NULL)            p = root;        q = p;        while(q != NULL && q->num != -1)        {            ans[q->num][0] = x - (len[q->num] - 1) * dir[k][0];            ans[q->num][1] = y - (len[q->num] - 1)* dir[k][1];            ans[q->num][2] = k;            q->num = -1;            q = q->fail;        }                x += dir[k][0];        y += dir[k][1];    }}