[AHOI 2013]差异

题意：求\(\sum_{1<=i<j<=n} len(T_i) + len(T_j) - 2*LCP(T_i,T_j)\)

思路：首先考虑把式子化简一下：

原式
= \(\sum_{1<=i<j<=n} len(T_i) + len(T_j) - 2 * \sum_{1<=i<j<=n}LCP(T_i,T_j)\)

= $\sum_{1<=i<j<=n} (n - i + 1) + (n - j + 1) - 2 * \sum_{1<=i<j<=n}LCP(T_i,T_j) $
= \(((n + 1) * n * (n - 1) / 2) - 2 * \sum_{1<=i<j<=n}LCP(T_i,T_j)\)

前面\(O(1)\)算一下，关键是后面求最长公共前缀，\(SAM\)跑一下即可，或者后缀树\(dp\)（懒得写了）。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 500005;
char s[maxn];
int cnt = 1;
int lst = 1;
int rt = 1;
int son[maxn << 1][26];
int fa[maxn << 1];
int dis[maxn << 1];
int siz[maxn << 1];
int _cnt;
struct edge {
	int to;
	int nxt;
}e[maxn << 2];
ll ans;
int head[maxn<<2];
inline void extend(int c) {
	int p = lst;
	int np = ++cnt;
	lst = np;
	siz[np] = 1;
	while(p && !son[p][c]) son[p][c] = np,p = fa[p];
	if(!p) {
		fa[np] = rt;
	}
	else {
		int q = son[p][c];
		if(dis[q] == dis[p] + 1) fa[np] = q;
		else {
			int nq = ++cnt;
			memcpy(son[nq],son[q],sizeof(son[nq]));
			fa[nq] = fa[q];
			dis[nq] = dis[p] + 1;
			fa[np] = fa[q] = nq;
			while(son[p][c] == q) {
				son[p][c] = nq;
				p = fa[p];
			}
		}
	}
}
inline void Add_edge(int u,int v) {
	e[++_cnt].to = v;
	e[_cnt].nxt = head[u];
	head[u] = _cnt;return;
}
inline void dfs(int x) {
	for(int i = head[x];i;i=e[i].nxt) {
		int y = e[i].to;
		dfs(y);
		ans += (ll)siz[x] * siz[y] * dis[x];
		siz[x] += siz[y];
	}
}
int main () {
	scanf("%s",s + 1);
	int len = strlen(s + 1);
	for(int i = len;i >= 1; --i) {
		extend(s[i] - 'a');
	}
	for(int i = 2;i <= cnt; ++i) {
		Add_edge(fa[i],i);
	}
	dfs(1);
	printf("%lld\n",(ll)(len - 1) * len * (len + 1) / 2 - ans * 2);
	return 0;
}