就是在建好的回文树上进行DFS计数。
0,0代表偶数
1,1代表奇数
然后不断往后暴力添加字符，判断是否存在后继新的回文串节点，计数相称即可。
1 #include <bits/stdc++.h> 2 const long long mod = 1e9+7; 3 const double ex = 1e-10; 4 #define inf 0x3f3f3f3f 5 using namespace std; 6 const int MAXN = 200022; 7 const int N = 26; 8 char A[200022],B[200022]; 9 struct Palindromic_Tree{ 10 int next[MAXN][N]; 11 int fail[MAXN]; 12 int cnt[MAXN]; 13 int num[MAXN]; 14 int len[MAXN]; 15 int S[MAXN]; 16 int last; 17 int n; 18 int p; 19 int newnode(int l){ 20 for (int i = 0; i < N; i++) next[p][i] = 0; 21 cnt[p] = 0; 22 num[p] = 0; 23 len[p] = l; 24 return p++; 25 } 26 void init(){ 27 p = 0; 28 newnode(0); 29 newnode(-1); 30 last = 0; 31 n = 0; 32 S[n] = -1; 33 fail[0] = 1; 34 } 35 int get_fail(int x){ 36 while (S[n - len[x] - 1] != S[n]) x=fail[x]; 37 return x; 38 } 39 void add(int c){ 40 c -= 'a'; 41 S[++n] = c; 42 int cur = get_fail(last); 43 if (!next[cur][c]){ 44 int now = newnode(len[cur] + 2); 45 fail[now] = next[get_fail(fail[cur])][c]; 46 next[cur][c] = now; 47 num[now] = num[fail[now]] + 1; 48 } 49 last = next[cur][c]; 50 cnt[last]++; 51 } 52 void count(){ 53 for (int i = p - 1; i>=0; --i) cnt[fail[i]] += cnt[i]; 54 } 55 }; 56 Palindromic_Tree a,b; 57 long long dfs(int x,int y){ 58 long long tans = 0; 59 for (int i = 0 ; i< 26 ; i++){ 60 if (a.next[x][i]&&b.next[y][i]){ 61 tans += 1LL*a.cnt[a.next[x][i]]*b.cnt[b.next[y][i]] + dfs(a.next[x][i],b.next[y][i]); 62 } 63 } 64 return tans; 65 } 66 int main() 67 { 68 int T; 69 scanf("%d",&T); 70 for (int cas = 1; cas <= T; cas++){ 71 scanf(" %s",A); 72 scanf(" %s",B); 73 a.init(); 74 b.init(); 75 int la = strlen(A); 76 int lb = strlen(B); 77 for (int i = 0 ; i < la ; i++){ 78 a.add(A[i]); 79 } 80 for (int i = 0 ; i < lb ; i++){ 81 b.add(B[i]); 82 } 83 a.count(); 84 b.count(); 85 long long ans = 0; 86 ans = dfs(0,0) + dfs(1,1); 87 printf("Case #%d: %lld ",cas,ans); 88 } 89 }