bzoj 3926

后缀自动机扩展到树形结构上。
先建出大的Trie，然后我们得到了一棵Trie树，对于树上的每个节点，保存一个后缀自动机从根走它代表的字符串后到达的节点，每次其儿子就从父亲的这个节点开始扩展。
  1 /**************************************************************
  2     Problem: 3926
  3     User: idy002
  4     Language: C++
  5     Result: Accepted
  6     Time:4320 ms
  7     Memory:460976 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cassert>
 12 #include <cstring>
 13 #include <algorithm>
 14 #define N 100010
 15 #define ST 3000100
 16 #define SS ST<<1
 17 using namespace std;
 18  
 19 typedef long long dnt;
 20  
 21 int qu[SS], stk[SS], bg, ed, top;
 22  
 23 struct Sam {
 24     int son[SS][10], val[SS], pnt[SS], ntot;
 25     Sam() { pnt[0] = -1; };
 26     int append( int p, int c ) {
 27         int np = ++ntot;
 28         val[np] = val[p]+1;
 29         while( p!=-1 && !son[p][c] ) 
 30             son[p][c]=np, p=pnt[p];
 31         if( p==-1 ) {
 32             pnt[np] = 0;
 33         } else {
 34             int q=son[p][c];
 35             if( val[q]==val[p]+1 ) {
 36                 pnt[np] = q;
 37             } else {
 38                 int nq = ++ntot;
 39                 memcpy( son[nq], son[q], sizeof(son[nq]) );
 40                 val[nq] = val[p]+1;
 41                 pnt[nq] = pnt[q];
 42                 pnt[q] = pnt[np] = nq;
 43                 while( p!=-1 && son[p][c]==q )
 44                     son[p][c]=nq, p=pnt[p];
 45             }
 46         }
 47         return np;
 48     }
 49     dnt count() {
 50         dnt rt = 0;
 51         for( int u=1; u<=ntot; u++ ) 
 52             rt += val[u]-val[pnt[u]];
 53         return rt;
 54     }
 55 }sam;
 56 struct Trie {
 57     int son[ST][10], last[ST], ntot;
 58     void insert( int *T ) {
 59         int u=0;
 60         for( int i=0; T[i]!=-1; i++ ) {
 61             int c=T[i];
 62             if( !(0<=T[i]&&T[i]<10) ) {
 63                 assert( T[i]>=0 && T[i]<=9 );
 64             }
 65             if( !son[u][c] ) son[u][c]=++ntot;
 66             u=son[u][c];
 67         }
 68     }
 69     void build() {
 70         last[0] = 0;
 71         qu[bg=ed=1] = 0;
 72         while( bg<=ed ) {
 73             int u=qu[bg++];
 74             for( int c=0; c<10; c++ ) {
 75                 int v=son[u][c];
 76                 if( !v ) continue;
 77                 last[v] = sam.append( last[u], c );
 78                 qu[++ed] = v;
 79             }
 80         }
 81     }
 82 }trie;
 83  
 84 int n, c;
 85 int head[N], wght[N], dest[N<<1], next[N<<1], etot;
 86 int dgr[N], fat[N];
 87  
 88 void adde( int u, int v ) {
 89     etot++;
 90     dest[etot] = v;
 91     next[etot] = head[u];
 92     head[u] = etot;
 93 }
 94 void bfs( int s ) {
 95     qu[bg=ed=1] = s;
 96     fat[s] = 0;
 97     while( bg<=ed ) {
 98         int u=qu[bg++];
 99         for( int t=head[u]; t; t=next[t] ) {
100             int v=dest[t];
101             if( v==fat[u] ) continue;
102             qu[++ed] = v;
103             fat[v] = u;
104         }
105     }
106     for( int t=1; t<=top; t++ ) {
107         int u=stk[t];
108         bg=1, ed=0;
109         while( u ) {
110             qu[++ed] = wght[u];
111             u=fat[u];
112         }
113         reverse( qu+1, qu+1+ed );
114         qu[++ed] = -1;
115         trie.insert( qu+1 );
116     }
117 }
118 int main() {
119     scanf( "%d%d", &n, &c );
120     for( int i=1; i<=n; i++ ) 
121         scanf( "%d", wght+i );
122     for( int i=1,u,v; i<n; i++ ) {
123         scanf( "%d%d", &u, &v );
124         adde( u, v );
125         adde( v, u );
126         dgr[u]++, dgr[v]++;
127     }
128     for( int u=1; u<=n; u++ ) 
129         if( dgr[u]<=1 ) stk[++top] = u;
130     for( int t=1; t<=top; t++ ) 
131         bfs(stk[t]);
132     trie.build();
133     printf( "%lld
", sam.count() );
134     return 0;
135 }
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