思路

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

const int N = 1000010;

int n;
int tr[N][26], f[N],idx;//f存下的是每一个点走过的次数
int  ne[N];
char str[N];
int id[210];
vector<int> v;//存下每一个入队的点
void insert(int x)
{
    int p = 0;
    for (int i = 0; str[i]; i ++ )
    {
        int t = str[i] - 'a';
        if (!tr[p][t]) tr[p][t] = ++ idx;
        p = tr[p][t];
        f[p] ++ ;
    }
    id[x] = p;
}

void build()
{
    queue<int> q;
    for (int i = 0; i < 26; i ++ )
        if (tr[0][i])
                q.push(tr[0][i]);

    while (q.size())
    {
        int t = q.front();
        q.pop();
        v.push_back(t);
        for (int i = 0; i < 26; i ++ )
        {
            int &p = tr[t][i];
            if (!p) p = tr[ne[t]][i];
            else
            {
                ne[p] = tr[ne[t]][i];
                q.push(p);
            }
        }
    }
}

int main()
{
    scanf("%d", &n);

    for (int i = 0; i < n; i ++ )
    {
        scanf("%s", str);
        insert(i);
    }

    build();

    for (int i = idx - 1; i >= 0; i -- ) f[ne[v[i]]] += f[v[i]];

    for (int i = 0; i < n; i ++ ) printf("%d\n", f[id[i]]);

    return 0;
}