其实是看后缀数组资料看到这个题目的,但是一眼反应显然后缀自动机,每次维护添加节点后的答案贡献即可,唯一不友好的一点是需要平衡树维护,这里因为复杂度不卡而且也不需要用到(ch)数组的遍历访问,我采用了(map)的写法。
其实是因为不会平衡树维护啦,如果有机会还是要学一下啊。。这个题可是没给不用平衡树的部分分啊(Qwq)(直接全体(1e9))
#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
struct Suffix_Auto {
int cnt, las; long long ans;
int fa[N], len[N];
map <int, int> ch[N];
Suffix_Auto () {
cnt = las = 1; ans = 0;
memset (fa, 0, sizeof (fa));
memset (len, 0, sizeof (len));
}
void extend (int c) {
int p = las, q = ++cnt; las = q;
len[q] = len[p] + 1;
while (p != 0 && ch[p][c] == 0) {
ch[p][c] = q;
p = fa[p];
}
ans += len[q];
if (p == 0) {
fa[q] = 1;
} else {
int x = ch[p][c];
if (len[x] == len[p] + 1) {
fa[q] = x;
ans -= len[x];
} else {
int y = ++cnt;
fa[y] = fa[x];
fa[x] = fa[q] = y;
len[y] = len[p] + 1;
ch[y] = ch[x];
while (p != 0 && ch[p][c] == x) {
ch[p][c] = y;
p = fa[p];
}
ans -= len[y];
}
}
cout << ans << endl;
}
}sam;
int n, num;
int main () {
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> num;
sam.extend (num);
}
}