题面
Sol
刚完品酒大会那道题后再看这道题发现这就是道\(SB\)题
后缀数组+并查集
按\(height\)从大到小做
\(height\)是两个相邻\(rank\)的后缀的\(LCP\)
从大到小，那么每次合并\(height\)的两边的集合，同时记录答案
两边集合两两配对的\(LCP\)一定就是这个\(height\)，乘法原理就可以了
这也算是一种套路吧
代码我常数大我最菜
# 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;
const int _(5e5 + 5);
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;
}
int n, s[_], sa[_], rk[_], tmp[_], t[_], height[_], id[_];
int fa[_], size[_];
ll ans;
char ss[_];
IL int Find(RG int x){
return x == fa[x] ? x : fa[x] = Find(fa[x]);
}
IL ll S(RG ll x){
return x * (x + 1) / 2;
}
IL int Cmp(RG int i, RG int j, RG int k){
return tmp[i] == tmp[j] && i + k <= n && j + k <= n && tmp[i + k] == tmp[j + k];
}
IL void Suffix_Sort(){
RG int m = 26;
for(RG int i = 1; i <= n; ++i) ++t[rk[i] = s[i]];
for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
for(RG int k = 1; k <= n; k <<= 1){
RG int l = 0;
for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;
for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;
for(RG int i = 1; i <= m; ++i) t[i] = 0;
for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];
for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];
swap(rk, tmp), rk[sa[1]] = l = 1;
for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
if(l >= n) break;
m = l;
}
for(RG int i = 1, h = 0; i <= n; ++i){
if(h) --h;
while(s[i + h] == s[sa[rk[i] - 1] + h]) ++h;
height[rk[i]] = h;
}
}
IL int _Cmp(RG int x, RG int y){
return height[x] > height[y];
}
int main(RG int argc, RG char* argv[]){
scanf(" %s", ss + 1), n = strlen(ss + 1);
for(RG int i = 1; i <= n; ++i){
s[i] = ss[i] - 'a' + 1, id[i] = fa[i] = i, size[i] = 1;
ans += S(n - i) + 1LL * (n - i) * (n - i + 1);
}
Suffix_Sort(), sort(id + 2, id + n + 1, _Cmp);
for(RG int i = 2; i <= n; ++i){
RG int x = Find(sa[id[i] - 1]), y = Find(sa[id[i]]);
ans -= 2LL * size[x] * size[y] * height[id[i]];
fa[x] = y, size[y] += size[x];
}
printf("%lld\n", ans);
return 0;
}