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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
", ans);
return 0;
}
后缀自动机也可以做
就是(parent)树上每个点的(len)的贡献，取决于它不同子树的(endpos(right))集合的乘积和
就可以求出每对前缀的最长公共后缀长度的和
那么把串翻转做就是题目要求的了
# 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(1e6 + 5);
int fa[maxn], trans[26][maxn], len[maxn], last = 1, tot = 1, n;
int size[maxn], t[maxn], id[maxn];
char s[maxn];
ll ans;
IL void Extend(RG int c){
RG int p = last, np = ++tot;
last = np, len[np] = len[p] + 1, size[np] = 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[p] + 1 == len[q]) 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[q] = fa[np] = nq;
while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
}
}
}
int main(RG int argc, RG char *argv[]){
scanf(" %s", s), n = strlen(s);
ans = (ll) (n - 1) * n * (n + 1) >> 1;
for(RG int i = n - 1; ~i; --i) Extend(s[i] - 'a');
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 p = id[i];
ans -= (ll) size[p] * size[fa[p]] * len[fa[p]] << 1;
size[fa[p]] += size[p];
}
printf("%lld
", ans);
return 0;
}