给定一个长度为 nn 的字符串 SS,令 TiT_i表示它从第 ii个字符开始的后缀,求:

1i<jnlen(Ti)+len(Tj)2lcp(Ti,Tj)\sum_{1 \le i < j \le n} len(T_i)+len(T_j)-2*lcp(T_i,T_j)

LOJ 2377

Solution

lenlen的部分直接算,后面的部分直接反着建SAM,在parent树上统计贡献即可

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
T sum = 0, fl = 1; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
return sum * fl;
}

inline void proc_status ()
{
ifstream t ("/proc/self/status");
cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int MAXN = 500000 + 100;

int N;
char S[MAXN];

namespace SAM
{
const int MAX_NODE = MAXN << 1;

int node_cnt = 1, last = 1;

struct info
{
int ch[27], fa, maxlen;
LL cnt;
} node[MAX_NODE];

vector <int> G[MAX_NODE];

inline int new_node (int pre)
{
node[++node_cnt].maxlen = node[pre].maxlen + 1;
return node_cnt;
}

inline void extend (int c)
{
int now = new_node (last), pre = last; last = now;
node[now].cnt = 1;

for (; pre && !node[pre].ch[c]; pre = node[pre].fa) node[pre].ch[c] = now;

if (!pre) node[now].fa = 1;
else
{
int x = node[pre].ch[c];
if (node[x].maxlen == node[pre].maxlen + 1) node[now].fa = x;
else
{
int y = ++node_cnt;
node[y] = node[x], node[y].maxlen = node[pre].maxlen + 1, node[y].cnt = 0;
node[x].fa = node[now].fa = y;

for (; node[pre].ch[c] == x; pre = node[pre].fa) node[pre].ch[c] = y;
}
}
}

inline void build (char *s, int n) { for (int i = 1; i <= n; ++i) extend (s[i] - 'a'); }

LL ans;

inline void dfs (int x)
{
LL sum = 0, done = node[x].cnt;

for (int i = 0; i < G[x].size(); ++i)
{
int y = G[x][i];
dfs (y);
sum += done * node[y].cnt;
done += node[y].cnt;
}

ans += sum * node[x].maxlen;
node[x].cnt = done;
}

inline LL get_ans ()
{
for (int i = 1; i <= node_cnt; ++i) G[node[i].fa].pb (i);
dfs (1);
return ans;
}
}

inline LL Calc (int l, int r) { return (LL)(l + r) * (r - l + 1) / 2ll; }

inline void Solve ()
{
reverse (S + 1, S + N + 1);
SAM :: build (S, N);

LL ans = 0;
for (int i = 1; i <= N; ++i) ans += (LL)Calc (1, N - i) + (LL)(N - i) * (N - i + 1);

printf("%lld\n", ans - 2ll * SAM:: get_ans ());
}

inline void Input ()
{
scanf("%s", S + 1);
N = strlen (S + 1);
}

int main()
{

#ifndef ONLINE_JUDGE
freopen("A.in", "r", stdin);
freopen("A.out", "w", stdout);
#endif

Input ();
Solve ();

return 0;
}