CF452E Three strings 题解

建广义 SAM，考虑统计 $e_{i,0/1/2}$ 表示 $i$ 点在每个串中分别出现多少次，

也就是说，对于 $T$ 串的每个前缀，将其对应节点的 link 树根链上每个点 $u$ 的 $e_{u,T}$ 加上 $1$。

可以将其树上差分为单点加，子树求和，最终可以得到每个 $i$ 点的 $e_{i,0/1/2}$。

考虑每个点的贡献，显然 $i$ 点会使 $(\text{len}(\text{link}(i)),\text{len}(i)]$ 的答案各加上 $e_{i,0}\times e_{i,1}\times e_{i,2}$。

#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
#define M 1000000007
using namespace std;
queue<int> q;
int T, n, o, z, e[600050][3], l[600050], f[600050], d[600050], c[600050][26];
long long Q[600050];
string u[600050];
void I(int x)
{
    int u = ++o, p = z;
    l[u] = l[z] + 1;
    while (p != -1 && !c[p][x])
        c[p][x] = u, p = f[p];
    if (p == -1)
        f[u] = 0;
    else
    {
        int q = c[p][x];
        if (l[p] + 1 == l[q])
            f[u] = q;
        else
        {
            int w = ++o;
            l[w] = l[p] + 1;
            f[w] = f[q];
            for (int i = 0; i < 26; ++i)
                c[w][i] = c[q][i];
            while (p != -1 && c[p][x] == q)
                c[p][x] = w, p = f[p];
            f[q] = f[u] = w;
        }
    }
    z = u;
}
int main()
{
    f[0] = -1;
    T = 3;
    for (int i = 0; i < T; ++i)
    {
        cin >> u[i];
        z = 0;
        for (auto j : u[i])
            I(j - 'a');
    }
    for (int i = 0; i < T; ++i)
        for (int j = 0, p = 0; j < u[i].size(); ++j)
            p = c[p][u[i][j] - 'a'], ++e[p][i];
    for (int i = 1; i <= o; ++i)
        ++d[f[i]];
    for (int i = 1; i <= o; ++i)
        if (!d[i])
            q.push(i);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        if (!u)
            continue;
        Q[l[f[u]] + 1] = (Q[l[f[u]] + 1] + 1ll * e[u][0] * e[u][1] % M * e[u][2] % M) % M;
        Q[l[u] + 1] = (Q[l[u] + 1] + M - 1ll * e[u][0] * e[u][1] % M * e[u][2] % M) % M;
        for (int i = 0; i < 3; ++i)
            e[f[u]][i] += e[u][i];
        if (!--d[f[u]])
            q.push(f[u]);
    }
    for (int i = 1; i <= (int)min({u[0].length(), u[1].length(), u[2].length()}); ++i)
        cout << (Q[i] = (Q[i] + Q[i - 1]) % M) << ' ';
    return 0;
}