Bzoj3277：3277

题面

传送门

Sol

广义\(sam\)
每个\(sam\)的状态开\(set\)记录属于哪些串
\(parent\)树上启发式合并\(set\)

然后每个串就在上面走，通过不停地跳\(parent\)树的父亲节点保证大于等于\(k\)，贡献就是\(len\)

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
 
IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}
 
const int maxn(1e5 + 5);
const int maxm(5e5 + 5);

int trans[26][maxm], fa[maxm], tot = 1, last, len[maxm];
int id[maxm], t[maxm], n, k, size[maxm];
set <int> ed[maxm];
set <int> :: iterator it;
string s[maxn];

IL void Extend(RG int c){
    RG int p = last, np = ++tot;
    len[last = np] = len[p] + 1;
    while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
    if(!p) fa[np] = 1;
    else{
        RG int q = trans[c][p];
        if(len[q] == len[p] + 1) fa[np] = q;
        else{
            RG int nq = ++tot;
            fa[nq] = fa[q], len[nq] = len[p] + 1;
            for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
            fa[np] = fa[q] = nq;
            while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
        }
    }
}

int main(RG int argc, RG char *argv[]){
    n = Input(), k = Input();
    for(RG int i = 1; i <= n; ++i){
        cin >> s[i]; last = 1;
        for(RG int j = 0, l = s[i].size(); j < l; ++j) Extend(s[i][j] - 'a');
    }
    for(RG int i = 1; i <= n; ++i)
        for(RG int j = 0, nw = 1, l = s[i].size(); j < l; ++j){
            nw = trans[s[i][j] - 'a'][nw];
            ed[nw].insert(i);
        }
    for(RG int i = 1; i <= tot; ++i) ++t[len[i]];
    for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
    for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
    for(RG int i = tot; i; --i){
        RG int x = id[i];
        size[x] = ed[x].size();
        if(size[x] > size[fa[x]]) swap(ed[x], ed[fa[x]]);
        for(it = ed[x].begin(); it != ed[x].end(); ++it)
            ed[fa[x]].insert(*it);
    }
    for(RG int i = 1; i <= n; ++i){
        RG ll ans = 0, nw = 1;
        for(RG int j = 0, l = s[i].size(); j < l; ++j){
            nw = trans[s[i][j] - 'a'][nw];
            while(nw != 1 && size[nw] < k) nw = fa[nw];
            if(size[nw] >= k) ans += len[nw];
        }
        printf("%lld ", ans);
    }
    return 0;
}