bzoj3238: [Ahoi2013]差异

Description



Input

一行，一个字符串S

Output

一行，一个整数，表示所求值

Sample Input

cacao

Sample Output

54

HINT

2<=N<=500000,S由小写英文字母组成

简单来说就是求原串中所有后缀两两最长公共前缀之和..

那么只要把原串反过来做SAM，把原问题变成所有前缀两两最长公共后缀之和

然后就要用到2015集训队论文中的性质5



那么题解就是显而易见的啊..

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define LL long long
using namespace std;
const LL Maxn = 1000010;
LL F[Maxn], d[Maxn], ch[Maxn][26], tot, now, size[Maxn], lca[Maxn];
char s[Maxn]; LL len;
LL Rsort[Maxn], rk[Maxn];
LL copy ( LL p, LL c ){
LL x = ++tot, y = ch[p][c];
d[x] = d[p]+1;
for ( LL i = 0; i < 26; i ++ ) ch[x][i] = ch[y][i];
F[x] = F[y]; F[y] = x;
while ( ~p && ch[p][c] == y ){ ch[p][c] = x; p = F[p]; }
return x;
}
void add ( LL c ){
LL p, o;
if ( p = ch[now][c] ){
if ( d[p] != d[now]+1 ) copy ( now, c );
now = ch[now][c];
}
else {
d[o=++tot] = d[now]+1; p = now; now = o;
while ( ~p && !ch[p][c] ){ ch[p][c] = o; p = F[p]; }
F[o] = ~p ? ( d[p]+1 == d[ch[p][c]] ? ch[p][c] : copy ( p, c ) ) : 0;
}
}
int main (){
LL i, j, k;
scanf ( "%s", s+1 );
len = strlen (s+1);
F[0] = -1; tot = now = 0;
for ( i = len; i >= 1; i -- ){ add (s[i]-'a'); size[now] = 1; }
for ( i = 1; i <= tot; i ++ ) Rsort[d[i]] ++;
for ( i = 1; i <= len; i ++ ) Rsort[i] += Rsort[i-1];
for ( i = tot; i >= 1; i -- ) rk[Rsort[d[i]]--] = i;
LL ans = 0;
for ( i = tot; i >= 1; i -- ){
lca[F[rk[i]]] += size[F[rk[i]]]*size[rk[i]];
size[F[rk[i]]] += size[rk[i]];
ans += lca[rk[i]]*d[rk[i]];
}
printf ( "%lld\n", (len-1)*len*(len+1)/2-2*ans );
return 0;
}