BZOJ 4516: [Sdoi2016]生成魔咒 [后缀自动机]

4516: [Sdoi2016]生成魔咒

题意：询问一个字符串每个前缀有多少不同的子串

---

做了一下SDOI2016R1D2，题好水啊随便AK

强行开map上SAM
每个状态的贡献就是(Max(s)-Min(s)+1)
插入的时候维护一下就行了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N=3e5+5, P=1e9+7;
inline int read() {
	char c=getchar(); int x=0, f=1;
	while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
	while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
	return x*f;
}

int n, s[N]; ll ans;
struct meow { map<int, int> ch; int par, val;}t[N];
int sz=1, root=1, last=1;
void extend(int c) {
	int p=last, np=++sz;
	t[np].val = t[p].val+1;
	for(; p && !t[p].ch[c]; p=t[p].par) t[p].ch[c]=np;
	if(!p) t[np].par = root;
	else {
		int q=t[p].ch[c];
		if(t[q].val == t[p].val+1) t[np].par=q;
		else {
			ans -= t[q].val - t[p].val;
			int nq=++sz; t[nq]=t[q]; t[nq].val = t[p].val+1;
			t[q].par = t[np].par = nq;
			ans += t[q].val - t[nq].val; ans ++;
			for(; p && t[p].ch[c]==q; p=t[p].par) t[p].ch[c] = nq;
		}
	}
	ans+=t[np].val - t[t[np].par].val;
	last=np;
}
int main() {
	//freopen("in","r",stdin);
	freopen("incantation.in","r",stdin);
	freopen("incantation.out","w",stdout);
	n=read();
	for(int i=1; i<=n; i++) {
		s[i]=read();
		extend(s[i]);
		printf("%lld
",ans);
	}
}