POJ 1204 AC自动机

题目就是给出了一个矩阵，由大写字母构成，然后让你查找某些单词在矩阵中出现的位置

出现的方式可能有8种，从某个位置往北连续的字符串，往东北，东........八个方向的只要有满足的 就可以

最后输出位置和方向，然后往北的输出时为A，东北的是B，依次顺时针类推

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-5#define MAXN 1111#define MAXM 222200#define INF 1000000000using namespace std;int e, cnt, L, C, W;char s[MAXN][MAXN];bool vis[MAXN];int ansx[MAXN], ansy[MAXN], len[MAXN];int loc[MAXN];int tx[8] = {-1, -1, 0, 1, 1, 1, 0, -1}, ty[8] = {0, 1, 1, 1, 0, -1, -1, -1};char dic[] = "ABCDEFGH";char tmp[MAXN];struct Trie{    int next[26];    int fail, id;    void init ()    {        memset(next, 0, sizeof(next));        fail = -1;        id = 0;    }} trie[MAXM];void make_trie (char *str, int id){    int i = 0, index, u = 0;    while(str[i])    {        index = str[i] - 'A';        if(!trie[u].next[index]) trie[u].next[index] = ++e;        u = trie[u].next[index];        i++;    }    trie[u].id = id;}int q[MAXM];void make_ac_automation(){    int i, j;    int h = 0,  t = 0;    q[t++] = 0;    while(h < t)    {        int u = q[h++];        for(i = 0; i < 26; i++)            if(trie[u].next[i])            {                int v = trie[u].next[i];                for(j = trie[u].fail; j != -1; j = trie[j].fail)                    if(trie[j].next[i])                    {                        trie[v].fail = trie[j].next[i];                        break;                    }                if(j == -1) trie[v].fail = 0;                q[t++] = v;            }    }}void match(int x, int y, int k){    int u = 0, index;    while(s[x][y])    {        index = s[x][y] - 'A';        while(!trie[u].next[index] && u != 0) u = trie[u].fail;        u = trie[u].next[index];        if(u == -1) u = 0;        int v = u;        while(v != 0 && trie[v].id != -1)        {            if(trie[v].id > 0 && !vis[trie[v].id])            {                cnt++;                vis[trie[v].id] = 1;                ansx[trie[v].id] = x - (len[trie[v].id] - 1) * tx[k];                ansy[trie[v].id] = y - (len[trie[v].id] - 1) * ty[k];                loc[trie[v].id] = k;                if(cnt == W) return;            }            trie[v].id = -1;            v = trie[v].fail;        }        x += tx[k], y += ty[k];    }}int main(){    for(int i = 0; i < MAXM; i++) trie[i].init();    scanf("%d%d%d", &L, &C, &W);    for(int i = 1; i <= L; i++) scanf("%s", s[i] + 1);    for(int i = 1; i <= W; i++)    {        scanf("%s", tmp);        make_trie(tmp, i);        len[i] = strlen(tmp);    }    make_ac_automation();    for(int i = 1; i <= C; i++)        match(L, i, 0);//A    for(int i = 1; i <= L; i++)        match(i, 1, 1);//B    for(int i = 1; i <= C; i++)        match(L, i, 1);//B    for(int i = 1; i <= L; i++)        match(i, 1, 2);//C    for(int i = 1; i <= C; i++)        match(1, i, 3);//D    for(int i = 1; i <= L; i++)        match(i, 1, 3);    for(int i = 1; i <= C; i++)        match(1, i, 4);//E    for(int i = 1; i <= L; i++)        match(i, C, 5);//F    for(int i = 1; i <= C; i++)        match(1, i, 5);    for(int i = 1; i <= L; i++)        match(i, C, 6);//G    for(int i = 1; i <= C; i++)        match(L, i, 7);//H    for(int i = 1; i <= L; i++)        match(i, C, 7);    for(int i = 1; i <= W; i++)        if(vis[i]) printf("%d %d %c\n", ansx[i] - 1, ansy[i] - 1, dic[loc[i]]);    return 0;}