bzoj4566 找相同字符

题意：给定两个字符串，从中各取一个子串使之相同，有多少种取法。允许本质相同。

解：建立广义后缀自动机，对于每个串，分别统计cnt，之后每个点的cnt乘起来。记得开long long

 #include <cstdio>
 #include <algorithm>
 #include <cstring>

 typedef long long LL;
 const int N = ;

 struct Edge {
     int nex, v;
 }edge[N << ]; int top;

 int tr[N][], len[N], fail[N], cnt[N][], vis[N];
 int tot = , last, turn, e[N];
 char s[N], str[N];

 inline void add(int x, int y) {
     top++;
     edge[top].v = y;
     edge[top].nex = e[x];
     e[x] = top;
     return;
 }

 inline int split(int p, int f) {
     int Q = tr[p][f], nQ = ++tot;
     len[nQ] = len[p] + ;
     fail[nQ] = fail[Q];
     fail[Q] = nQ;
     memcpy(tr[nQ], tr[Q], sizeof(tr[Q]));
     while(tr[p][f] == Q) {
         tr[p][f] = nQ;
         p = fail[p];
     }
     return nQ;
 }

 inline int insert(int p, char c) {
     int f = c - 'a';
     if(tr[p][f]) {
         int Q = tr[p][f];
         if(len[Q] == len[p] + ) {
             cnt[Q][turn] = ;
             return Q;
         }
         int t = split(p, f);
         cnt[t][turn] = ;
         return t;
     }
     int np = ++tot;
     len[np] = len[p] + ;
     cnt[np][turn] = ;
     while(p && !tr[p][f]) {
         tr[p][f] = np;
         p = fail[p];
     }
     if(!p) {
         fail[np] = ;
     }
     else {
         int Q = tr[p][f];
         if(len[Q] == len[p] + ) {
             fail[np] = Q;
         }
         else {
             fail[np] = split(p, f);
         }
     }
     return np;
 }

 void DFS(int x) {
     for(int i = e[x]; i; i = edge[i].nex) {
         int y = edge[i].v;
         DFS(y);
         cnt[x][] += cnt[y][];
         cnt[x][] += cnt[y][];
     }
     return;
 }

 int main() {
     scanf("%s%s", s, str);
     int n = strlen(s), last = ;
     for(int i = ; i < n; i++) {
         last = insert(last, s[i]);
     }
     n = strlen(str);
     last = turn = ;
     for(int i = ; i < n; i++) {
         last = insert(last, str[i]);
     }
     for(int i = ; i <= tot; i++) {
         add(fail[i], i);
     }
     DFS();
     int p = ;
     LL ans = ;
     for(int i = ; i <= tot; i++) {
         ans += 1ll * cnt[i][] * cnt[i][] * (len[i] - len[fail[i]]);
     }

     printf("%lld", ans);
     return ;
 }

AC代码