POJ 1204Word Puzzles解题报告

确实是一道字符串好题，用AC自动机做的，不过思路比较简单，并且AC自动机还要加一些优化，要找的是串是否出现在了模式串中，这样有很多位置不用去比较，比如说，要考虑一行的时候，这一行的子串就不用再次验证了，因为出现在了这一行的一部分中一定会出现在整个这一行中的，这样能减少处理时间

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 1005
struct node
{
    bool istail;
    node *next[26],*fail;
    int id;
}qq[100005],*que[100005];
node *rt = new node();
char key[N];
int len[N];
int dix[8][2] = {-1,0,-1,1,0,1,1,1,1,0,1,-1,0,-1,-1,-1};
char res[9] = {"ABCDEFGH"};
struct pos
{
    int x,y,dis;
};
pos ans[N];
char mes[N][N];
int k;
void insert(char *str,int id)
{
    int i;
    int temp;
    node *p = rt;
    for(i = 0;str[i]; i++)
    {
        temp = str[i] - 'A';
        if(p->next[temp] == NULL)
        {
            qq[k].istail = false;
            qq[k].fail = NULL;
            memset(qq[k].next,NULL,sizeof(qq[k].next));
            p->next[temp] = &qq[k++];
        }
        p = p->next[temp];
    }
    p->istail = true;
    p->id = id;
}
void ac()
{
    int f = 0,r = 0;
    int i;
    node *p,*temp;
    rt->fail = NULL;
    que[r++] = rt;
    while(f < r)
    {
        temp = que[f++];
        for(i = 0;i < 26;i++)
        {
            if(temp->next[i] == NULL)
                continue;
            if(temp == rt)
                temp->next[i]->fail = rt;
            else
            {
                for(p = temp->fail; p != NULL; p = p->fail)
                {
                    if(p->next[i] != NULL)
                    {
                        temp->next[i]->fail = p->next[i];
                        break;
                    }
                }
                if(p == NULL)
                {
                    temp->next[i]->fail = rt;
                }
            }
            que[r++] = temp->next[i];
        }
    }
}
void query(int x,int y,int i)
{
    int tem;
    int nx = x,ny = y;
    node *p,*q;
    p = rt;
    for(;mes[x][y];x += dix[i][0],y += dix[i][1])
    {
        tem = mes[x][y] - 'A';
        while(p->next[tem] == NULL &&p != rt)
            p = p->fail;
        p = p->next[tem];
        if(p == NULL)
            p = rt;
        q = p;
        while(q != rt)
        {
            if(q->istail)
            {
                q->istail = false;
                ans[q->id].x = x - dix[i][0] * (len[q->id] - 1) - 1;
                ans[q->id].y = y - dix[i][1] * (len[q->id] - 1) - 1;
                ans[q->id].dis = i;
            }
            q = q->fail;
        }
    }
}
int main()
{
    int r,c,m;
    int i,j;
    for(i = 0;i < 26; i++)
        rt->next[i] = NULL;
    memset(mes,0,sizeof(mes));
    scanf("%d%d%d",&r,&c,&m);
    for(i = 1;i <= r; i++)
        scanf("%s",mes[i]+1);
    for(i = 0;i < m;i++)
    {
        scanf("%s",key);
        len[i] = strlen(key);
        insert(key,i);
    }
    ac();
    for(i = 1;i <= r; i++)
    {
        query(i,1,1),query(i,1,2),query(i,1,3);
        query(i,c,5),query(i,c,6),query(i,c,7);
    }
    for(i = 1;i <= c; i++)
    {
        query(1,i,3),query(1,i,4),query(1,i,5);
        query(r,i,1),query(r,i,0),query(r,i,7);
    }
    for(i = 0;i < m;i++)
        printf("%d %d %c\n",ans[i].x,ans[i].y,res[ans[i].dis]);
    return 0;
}