本文主要是介绍【codeforces】163E. e-Government AC自动机+树状数组，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
传送门：【codeforces】163E. e-Government
题目分析：感觉到现在再做类似题目已经感觉很水了= =。。。这题也就是构建了fail指针树以后树状数组维护就好了。10^6个字母的意思就是说我们可以随便搞。。。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;typedef long long LL ;#define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define For( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )const int MAXN = 1000005 ;
const int MAXE = 1000005 ;struct Edge {int v , n ;Edge () {}Edge ( int var , int next ) : v ( var ) , n ( next ) {}
} ;struct ac_automaton {int next[MAXN][26] ;int fail[MAXN] ;int word[MAXN] ;int root ;int cur ;int Q[MAXN] ;int head ;int tail ;int T[MAXN] ;int in[MAXN] ;int ou[MAXN] ;int dfs_clock ;Edge E[MAXE] ;int H[MAXN] , cntE ;bool a[MAXN] ;int newnode () {rep ( i , 0 , 26 ) next[cur][i] = -1 ;return cur ++ ;}void init () {cur = 0 ;cntE = 0 ;dfs_clock = 0 ;clr ( a , 0 ) ;clr ( T , 0 ) ;clr ( H , -1 ) ;root = newnode () ;}void addedge ( int u , int v ) {E[cntE] = Edge ( v , H[u] ) ;H[u] = cntE ++ ;}void insert ( char buf[] , int idx ) {int now = root ;for ( int i = 0 ; buf[i] ; ++ i ) {int x = buf[i] - 'a' ;if ( next[now][x] == -1 ) next[now][x] = newnode () ;now = next[now][x] ;}word[idx] = now ;}void build () {head = tail = 0 ;fail[root] = root ;rep ( i , 0 , 26 ) {if ( ~next[root][i] ) {fail[next[root][i]] = root ;Q[tail ++] = next[root][i] ;} else next[root][i] = root ;}while ( head != tail ) {int now = Q[head ++] ;rep ( i , 0 , 26 ) {if ( ~next[now][i] ) {fail[next[now][i]] = next[fail[now]][i] ;Q[tail ++] = next[now][i] ;} else next[now][i] = next[fail[now]][i] ;}}rep ( i , 1 , cur ) addedge ( fail[i] , i ) ;}void dfs ( int u ) {in[u] = ++ dfs_clock ;for ( int i = H[u] ; ~i ; i = E[i].n ) dfs ( E[i].v ) ;ou[u] = dfs_clock ;}int sum ( int x , int ans = 0 ) {for ( int i = x ; i <= dfs_clock ; i += i & -i ) ans += T[i] ;return ans ;}void add ( int x , int v ) {for ( int i = x ; i ; i -= i & -i ) T[i] += v ;}void query ( char buf[] ) {int now = root ;LL ans = 0 ;for ( int i = 0 ; buf[i] ; ++ i ) {now = next[now][buf[i] - 'a'] ;//printf ( "%d %d %d\n" , ou[now] , in[now] , sum ( in[now] ) ) ;ans += sum ( in[now] ) ;}printf ( "%I64d\n" , ans ) ;}void modify ( int x , int sign ) {int now = word[x] ;if ( a[x] == 0 && sign == -1 || a[x] == 1 && sign == 1 ) return ;if ( a[x] == 0 ) {a[x] = 1 ;add ( ou[now] , 1 ) ;add ( in[now] - 1 , -1 ) ;} else {a[x] = 0 ;add ( ou[now] , -1 ) ;add ( in[now] - 1 , 1 ) ;}}
} ;ac_automaton ac ;
char buf[MAXN] ;
int n , k ;void solve () {char op ;int x ;ac.init () ;For ( i , 1 , k ) {scanf ( "%s" , buf ) ;ac.insert ( buf , i ) ;}ac.build () ;ac.dfs ( ac.root ) ;For ( i , 1 , k ) ac.modify ( i , 1 ) ;while ( n -- ) {scanf ( " %c" , &op ) ;if ( op == '?' ) {scanf ( "%s" , buf ) ;ac.query ( buf ) ;} else {scanf ( "%d" , &x ) ;ac.modify ( x , op == '+' ? 1 : -1 ) ;}}
}int main () {while ( ~scanf ( "%d%d" , &n , &k ) ) solve () ;return 0 ;
}这篇关于【codeforces】163E. e-Government AC自动机+树状数组的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
