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
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 5 typedef long long LL; 6 const int N = 800010; 7 8 struct Edge { 9 int nex, v; 10 }edge[N << 1]; int top; 11 12 int tr[N][26], len[N], fail[N], cnt[N][2], vis[N]; 13 int tot = 1, last, turn, e[N]; 14 char s[N], str[N]; 15 16 inline void add(int x, int y) { 17 top++; 18 edge[top].v = y; 19 edge[top].nex = e[x]; 20 e[x] = top; 21 return; 22 } 23 24 inline int split(int p, int f) { 25 int Q = tr[p][f], nQ = ++tot; 26 len[nQ] = len[p] + 1; 27 fail[nQ] = fail[Q]; 28 fail[Q] = nQ; 29 memcpy(tr[nQ], tr[Q], sizeof(tr[Q])); 30 while(tr[p][f] == Q) { 31 tr[p][f] = nQ; 32 p = fail[p]; 33 } 34 return nQ; 35 } 36 37 inline int insert(int p, char c) { 38 int f = c - 'a'; 39 if(tr[p][f]) { 40 int Q = tr[p][f]; 41 if(len[Q] == len[p] + 1) { 42 cnt[Q][turn] = 1; 43 return Q; 44 } 45 int t = split(p, f); 46 cnt[t][turn] = 1; 47 return t; 48 } 49 int np = ++tot; 50 len[np] = len[p] + 1; 51 cnt[np][turn] = 1; 52 while(p && !tr[p][f]) { 53 tr[p][f] = np; 54 p = fail[p]; 55 } 56 if(!p) { 57 fail[np] = 1; 58 } 59 else { 60 int Q = tr[p][f]; 61 if(len[Q] == len[p] + 1) { 62 fail[np] = Q; 63 } 64 else { 65 fail[np] = split(p, f); 66 } 67 } 68 return np; 69 } 70 71 void DFS(int x) { 72 for(int i = e[x]; i; i = edge[i].nex) { 73 int y = edge[i].v; 74 DFS(y); 75 cnt[x][0] += cnt[y][0]; 76 cnt[x][1] += cnt[y][1]; 77 } 78 return; 79 } 80 81 int main() { 82 scanf("%s%s", s, str); 83 int n = strlen(s), last = 1; 84 for(int i = 0; i < n; i++) { 85 last = insert(last, s[i]); 86 } 87 n = strlen(str); 88 last = turn = 1; 89 for(int i = 0; i < n; i++) { 90 last = insert(last, str[i]); 91 } 92 for(int i = 2; i <= tot; i++) { 93 add(fail[i], i); 94 } 95 DFS(1); 96 int p = 1; 97 LL ans = 0; 98 for(int i = 2; i <= tot; i++) { 99 ans += 1ll * cnt[i][0] * cnt[i][1] * (len[i] - len[fail[i]]); 100 } 101 102 printf("%lld", ans); 103 return 0; 104 }