串
n个字符串,对于每个串,有多少个子串至少存在于k个字符串中
right集合用set存出现的串标号。由于广义sam有废点(我这种写法),废点对parent树有影响,所以不能基排,直接建树跑dfs更新right集合,考虑启发式合并。之后每个串都扔进去跑一遍就行了。不需要考虑在sam中的出现次数,直接将整条fail边的长度贡献给答案。
#include<bits/stdc++.h>
using namespace std;
#define CLR(A, X) memset(A, X, sizeof(A))
typedef long long LL;
const int N = 2e5+10;
struct SAM {
int last, sz;
int ch[N][26], len[N], f[N], c[N], q[N], sum[N]; // N两倍大小
set<int> ed[N];
void init(int n) {
CLR(ch[1], 0);
for(int i = 0; i <= n; i++) {
c[i] = 0;
ed[i].clear();
}
last = sz = 1;
f[1] = len[1] = 0;
}
void Insert(int c, int id) {
int p = last, np = last = ++sz;
CLR(ch[np], 0);
len[np] = len[p] + 1; ed[np].insert(id);
while(!ch[p][c] && p) ch[p][c] = np, p = f[p];
if(!p) f[np] = 1;
else {
int q = ch[p][c];
if(len[q] == len[p]+1) f[np] = q;
else {
int nq = ++sz;
len[nq] = len[p]+1;
memcpy(ch[nq], ch[q], sizeof(ch[q]));
f[nq] = f[q]; f[np] = f[q] = nq;
while(p && ch[p][c]==q) ch[p][c] = nq, p = f[p];
}
}
}
}A;
void comb(set<int> &A, set<int> &B) {
if(B.size() > A.size()) swap(A, B);
for(set<int>::iterator it = B.begin(); it != B.end(); it++) A.insert(*it);
}
vector<int> G[N];
void dfs(int u) {
for(int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
dfs(v);
comb(A.ed[u], A.ed[v]);
}
A.sum[u] = A.ed[u].size();
}
char ss[N];
char *s[N];
int main() {
int n, K;
scanf("%d%d", &n, &K);
s[0] = ss;
A.init(2*n);
for(int i = 0; i < n; i++) {
scanf("%s", s[i]);
int len = strlen(s[i]);
s[i+1] = s[i]+len+1;
A.last = 1;
for(int j = 0; j < len; j++) {
A.Insert(s[i][j]-'a', i);
}
}
for(int i = 1; i <= A.sz; i++) {
G[A.f[i]].push_back(i);
}
dfs(1);
for(int i = 0; i < n; i++) {
int len = strlen(s[i]), u = 1;
LL ans = 0;
for(int j = 0; j < len; j++) {
u = A.ch[u][s[i][j]-'a'];
while(u!=1 && A.sum[u]<K) u = A.f[u];
ans += A.len[u];
}
printf("%lld%c", ans, i==n-1?'\n':' ');
}
return 0;
}