bzoj 4327: JSOI2012 玄武密码

听说这题不公开.. 那就不贴题意了

一眼看上去还以为是exkmp的裸题.. 看了数据范围，呵呵..

多串匹配嘛.. 就用AC自动机咯，而且每个点最多也就只有$4$个孩子

用原串在AC自动机上走，碰到的和fail到的都是可以到达的字符串，每个点标记记录一下，这个时间复杂度是$O(n)$的

然后再每个串走AC自动机，找到最远的被标记的点就是答案了..

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 using namespace std;
 const int Maxn = ;
 const int Maxm = ;
 struct trie {
     int child[], fl, fail;
 }tr[Maxm]; int tot, rt;
 char s[Maxm], st[Maxn][]; int n, stl[Maxn];
 int m;
 int get ( char c ){
     if ( c == 'N' ) return ;
     if ( c == 'W' ) return ;
     if ( c == 'E' ) return ;
     return ;
 }
 void bulid ( int &now, int l, int x ){
     if ( !now ) now = ++tot;
     if ( l == stl[x]+ ) return;
     bulid ( tr[now].child[get(st[x][l])], l+, x );
 }
 void bulid_ac (){
     queue <int> q;
     q.push (rt);
     while ( !q.empty () ){
         int x = q.front (); q.pop ();
         for ( int i = ; i < ; i ++ ){
             int y = tr[x].child[i];
             if ( !y ){
                 if ( x == rt ) tr[x].child[i] = x;
                 else tr[x].child[i] = tr[tr[x].fail].child[i];
             }
             else {
                 if ( x == rt ) tr[y].fail = x;
                 else tr[y].fail = tr[tr[x].fail].child[i];
             }
             if ( y >  ) q.push (y);
         }
     }
 }
 int find ( int now, int l, int x ){
     if ( l == stl[x]+ ) return stl[x];
     if ( tr[tr[now].child[get(st[x][l])]].fl ==  ) return l-;
     return find ( tr[now].child[get(st[x][l])], l+, x );
 }
 int main (){
     int i, j, k;
     scanf ( "%d%d", &n, &m );
     scanf ( "%s", s+ );
     for ( i = ; i <= m; i ++ ){
         scanf ( "%s", st[i]+ );
         stl[i] = strlen (st[i]+);
         bulid ( rt, , i );
     }
     bulid_ac ();
     int now = rt;
     for ( i = ; i <= n; i ++ ){
         now = tr[now].child[get(s[i])];
         int x = now;
         while ( tr[x].fl ==  && x ){
             tr[x].fl = ;
             x = tr[x].fail;
         }
     }
     for ( i = ; i <= m; i ++ ){
         printf ( "%d\n", find ( rt, , i ) );
     }
     return ;
 }