本文主要是介绍【HDU】4117 GRE Words AC自动机+线段树优化DP，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：【HDU】4117 GRE Words
题目分析：水不了啊狸的打字机就来水这题了= =。。。
首先建立ac自动机，然后用fail指针的反向关系建边，构造fail指针树。fail指针树中每个结点u表示的串都是其子节点v的后缀（同时该后缀是所有串中最长的）。对fail指针树dfs一次得到时间戳，当要求以串i结尾的最大价值，首先我们需要知道以串i的子串j结尾的最大价值val。因为在树中我们有关系如上所述，所以我们可以对i串的每个字符xi（对应树上一个结点）找到其在树上的价值最大的后缀，然后取最大加上串i本身的价值，即要求的val。然后我们对i的子树【in[i] , ou[i]】更新，即用val更新区间【in[i] , ou[i]】。查询的时候便是单点查询，查询in[xi]（xi为属于串i的每个字符）上的最大值。
我语文体育老师教的。
代码如下：
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;#pragma comment(linker, "/STACK:16777216")
#define rep( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define root 1 , 1 , ac.cur
#define mid ( ( l + r ) >> 1 )const int MAXN = 300005 ;
const int MAXE = 1000005 ;
const int INF = 0x3f3f3f3f ;struct ac_automaton {int next[MAXN][26] ;int fail[MAXN] ;int ac_root ;int cur ;int Q[MAXN] , head , tail ;int newnode () {rep ( i , 0 , 26 ) next[cur][i] = -1 ;return cur ++ ;}void clear () {cur = 0 ;ac_root = newnode () ;}inline int encode ( char c ) {return c - 'a' ;}void insert ( char* , int ) ;void build () {head = tail = 0 ;fail[ac_root] = ac_root ;rep ( i , 0 , 26 ) {if ( ~next[ac_root][i] ) {fail[next[ac_root][i]] = ac_root ;Q[tail ++] = next[ac_root][i] ;} else next[ac_root][i] = ac_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] ;}}}
} ;struct Edge {int v , n ;Edge () {}Edge ( int v , int n ) : v ( v ) , n ( n ) {}
} ;ac_automaton ac ;
Edge E[MAXE] ;
int lazy[MAXN << 2] ;
int maxv[MAXN << 2] ;
int H[MAXN] , cntE ;
int N[MAXN] , cntN ;
int d[MAXN] ;
int word[MAXN] ;
int val[MAXN] ;
int in[MAXN] , ou[MAXN] , dfs_clock ;
char buf[MAXN] ;
int n ;void clear () {cntE = 0 ;dfs_clock = 0 ;clr ( N , -1 ) ;clr ( H , -1 ) ;clr ( lazy , 0 ) ;clr ( maxv , 0 ) ;
}void addedge ( int u , int v , int H[] ) {E[cntE] = Edge ( v , H[u] ) ;H[u] = cntE ++ ;
}void ac_automaton :: insert ( char buf[] , int idx ) {int now = ac_root ;for ( int i = 0 ; buf[i] ; ++ i ) {int index = encode ( buf[i] ) ;if ( next[now][index] == -1 ) next[now][index] = newnode () ;now = next[now][index] ;addedge ( idx , now , N ) ;}word[idx] = now ;
}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 ;
}void push_down ( int o ) {if ( lazy[o] ) {lazy[ls] = max ( lazy[ls] , lazy[o] ) ;lazy[rs] = max ( lazy[rs] , lazy[o] ) ;maxv[ls] = max ( maxv[ls] , lazy[o] ) ;maxv[rs] = max ( maxv[rs] , lazy[o] ) ;lazy[o] = 0 ;}
}void push_up ( int o ) {maxv[o] = max ( maxv[ls] , maxv[rs] ) ;
}void update ( int L , int R , int v , int o , int l , int r ) {if ( l == r ) {lazy[o] = max ( lazy[o] , v ) ;maxv[o] = max ( maxv[o] , v ) ;return ;}push_down ( o ) ;int m = mid ;if ( L <= m ) update ( L , R , v , lson ) ;if ( m < R ) update ( L , R , v , rson ) ;push_up ( o ) ;
}int query ( int x , int o , int l , int r ) {if ( l == r ) return maxv[o] ;push_down ( o ) ;int m = mid ;if ( x <= m ) return query ( x , lson ) ;else return query ( x , rson ) ;
}void solve () {int x ;clear () ;ac.clear () ;scanf ( "%d" , &n ) ;For ( i , 1 , n ) {scanf ( "%s%d" , buf , &val[i] ) ;ac.insert ( buf , i ) ;}ac.build () ;rep ( i , 1 , ac.cur ) addedge ( ac.fail[i] , i , H ) ;dfs ( ac.ac_root ) ;int ans = 0 ;For ( u , 1 , n ) {if ( val[u] < 0 ) continue ;int tmp = 0 ;for ( int i = N[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;tmp = max ( tmp , query ( in[v] , root ) ) ;}ans = max ( tmp + val[u] , ans ) ;update ( in[word[u]] , ou[word[u]] , tmp + val[u] , root ) ;}printf ( "%d\n" , ans ) ;
}int main () {int T , cas = 0 ;scanf ( "%d" , &T ) ;while ( T -- ) {printf ( "Case #%d: " , ++ cas ) ;solve () ;}return 0 ;
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