注意事项：

  1. 函数 build 中 i-len[]-1 可能会超出字符串的下标范围，到达下标 0，需要特判。
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int cntnd, rootodd, rooteve;
int son[MAXlenb + 10][26 + 2], fail[MAXlenb + 10], len[MAXlenb + 10];
int cntep[MAXlenb + 10];
void build(int lenb, char *b) {
    cntnd = 2;
    rootodd = 1; len[rootodd] = -1;
    rooteve = 2; fail[rooteve] = rootodd; len[rooteve] = 0;
    int j = rootodd;
    for (int i = 1; i <= lenb; ++i) {
        while (i - len[j] - 1 == 0 || b[i - len[j] - 1] != b[i]) j = fail[j];
        if (son[j][b[i]] == 0) {
            int nwj = son[j][b[i]] = ++cntnd;
            len[nwj] = len[j] + 2;
            int k;
            if (j == rootodd) {
                k = rooteve;
            } else {
                k = fail[j];
                while (i - len[k] - 1 == 0 || b[i - len[k] - 1] != b[i]) k = fail[k];
                k = son[k][b[i]];
            }
            fail[nwj] = k;
        }
        j = son[j][b[i]];
        ++cntep[j];
    }
}

Luogu P3649 [APIO2014] 回文串

代码：

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXlenb = 3e5;

template <typename T>
inline void read(T &a) {
    char c;for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar());bool f = c == '-';T x = f ? 0 : (c ^ '0');for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) {x = x * 10 + (c ^ '0');}a = f ? -x : x;
}
template <typename T, typename ...Argv>
inline void read(T &a, Argv &...argv) {
    read(a), read(argv...);
}

int head[MAXlenb + 10], cntnex, nex[MAXlenb + 10], to[MAXlenb + 10];
inline void connect(int u, int v) {
    nex[++cntnex] = head[u];
    head[u] = cntnex;
    to[cntnex] = v;
}

int cntnd, rootodd, rooteve;
int son[MAXlenb + 10][26 + 2], fail[MAXlenb + 10], len[MAXlenb + 10];
int cntep[MAXlenb + 10];
void build(int lenb, char *b) {
    cntnd = 2;
    rootodd = 1; len[rootodd] = -1;
    rooteve = 2; fail[rooteve] = rootodd; len[rooteve] = 0;
    int j = rootodd;
    for (int i = 1; i <= lenb; ++i) {
        while (i - len[j] - 1 == 0 || b[i - len[j] - 1] != b[i]) j = fail[j];
        if (son[j][b[i]] == 0) {
            int nwj = son[j][b[i]] = ++cntnd;
            len[nwj] = len[j] + 2;
            int k;
            if (j == rootodd) {
                k = rooteve;
            } else {
                k = fail[j];
                while (i - len[k] - 1 == 0 || b[i - len[k] - 1] != b[i]) k = fail[k];
                k = son[k][b[i]];
            }
            fail[nwj] = k;
        }
        j = son[j][b[i]];
        ++cntep[j];
    }
}
void buildfailtree() {
    for (int i = 1; i <= cntnd; ++i) {
        if (i == rootodd) continue;
        connect(fail[i], i);
    }
}
ll ans;
void dfs(int cur) {
    for (int i = head[cur]; i; i = nex[i]) {
        dfs(to[i]);
        cntep[cur] += cntep[to[i]];
    }
    ans = max(ans, (ll)len[cur] * cntep[cur]);
}

int lenb; char b[MAXlenb + 10];
signed main() {
    scanf("%s", b + 1);
    lenb = strlen(b + 1);
    for (int i = 1; i <= lenb; ++i) b[i] -= 'a';
    build(lenb, b);
    buildfailtree();
    dfs(rootodd);
    printf("%lld\n", ans);
    return 0;
}