生成魔咒 LibreOJ - 2033

sam上dp。可以将trans[s,c]看成是边，整个sam就是一个dag图。我们令dp[u]表示u结尾的，不同字串个数有多少即可。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
struct sam {
	int len[N << 1];
	ll dp[N << 1];
	vector<map<int, int> >trans;
	vector<int>par;
	int last; int tot;
	int sz[N << 1];
	void init(int l) {
		trans.resize(l + 1, map<int, int>());
		par.resize(l + 1, 0);
		dp[1] = 1;
		last = 1; tot = 1;
	}
	ll ins(int c) {
		int pre = last; int now = last = ++tot;
		sz[now] = 1;
		len[now] = len[pre] + 1;
		for (; pre && !trans[pre][c]; pre = par[pre])
		{
			//int t = trans[pre][c];
			trans[pre][c] = now; dp[now] += dp[pre];
		}
		if (pre == 0) { par[now] = 1; return dp[now]; }
		int ano = trans[pre][c];
		if (len[ano] == len[pre] + 1) { par[now] = ano;}
		else {
			int nnow = ++tot;
			trans[nnow] = trans[ano];
			len[nnow] = len[pre] + 1;
			par[nnow] = par[ano];
			par[ano] = par[now] = nnow;
			for (; pre&&trans[pre][c] == ano; pre = par[pre])
			{
				dp[nnow] += dp[pre];
				trans[pre][c] = nnow;
			}
			dp[ano] -= dp[nnow];
		}
		return dp[now];
	}
}S;
int n;
int x[N];
int main()
{
	n = read();
	S.init(2 * n);
	ll ans = 0;
	upd(i, 1, n)
	{
		x[i] = read(); ans += S.ins(x[i]);
		printf("%lld\n",ans);
	}
	return 0;
}