【LOJ】 #2033. 「SDOI2016」生成魔咒
题解
就是字符集较大需要离散化和建边表的后缀自动机水题
每次会加入i个新的串,其中重复的就是i的父亲节点所在节点的长度,减掉即可
代码
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 1000005
//#define ivorysi
#define pb push_back
#define mo 1000007
#define pii pair<int,int>
#define mp make_pair
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 - '0' + c;
c = getchar();
}
res = res * f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
struct node {
int to,next,v;
}E[MAXN * 4];
int head[MAXN * 2],sumE,rt,last,Ncnt,par[MAXN * 2],len[MAXN * 2];
int find_Edge(int u,int v) {
for(int i = head[u] ; i ; i = E[i].next) {
if(E[i].v == v) return E[i].to;
}
return 0;
}
void add(int u,int v,int c) {
E[++sumE].to = v;E[sumE].v = c;E[sumE].next = head[u];
head[u] = sumE;
}
void build_SAM(int L,int v) {
int nowp = ++Ncnt,p;
len[nowp] = L;
for(p = last ; p && !find_Edge(p,v) ; p = par[p]) {
add(p,nowp,v);
}
if(!p) par[nowp] = rt;
else {
int q = find_Edge(p,v);
if(len[q] == len[p] + 1) par[nowp] = q;
else {
int copyq = ++Ncnt;
len[copyq] = len[p] + 1;par[copyq] = par[q];
for(int i = head[q] ; i ; i = E[i].next) {
add(copyq,E[i].to,E[i].v);
}
par[q] = copyq;par[nowp] = copyq;
for( ; p && find_Edge(p,v) == q ; p = par[p]) {
for(int i = head[p] ; i ; i = E[i].next) {
if(E[i].v == v) {E[i].to = copyq;break;}
}
}
}
}
last = nowp;
}
int N,a[MAXN],num[MAXN],tot;
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(a[i]);num[i] = a[i];
}
sort(num + 1,num + N + 1);
int tot = unique(num + 1,num + N + 1) - num - 1;
for(int i = 1 ; i <= N ; ++i) {
a[i] = lower_bound(num + 1,num + tot + 1,a[i]) - num;
}
rt = last = ++Ncnt;
int64 ans = 0;
for(int i = 1 ; i <= N ; ++i) {
build_SAM(i,a[i]);
ans += i - len[par[last]];
out(ans);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}