Distinct Substrings SPOJ - DISUBSTR

求不同字串的个数

 

 //刚开始个数为：n+(n-1)+.....+2+1 即（n+1）*n/2

//重复字串个数为height数组和

//sa[i]表示排名为i是从sa[i]位置开始的后缀
//rank[i]表示起始位置的下标为i的后缀的排名
// height[i]表示排名相邻的两个后缀的最长公共前缀长度
//https://blog.youkuaiyun.com/a1035719430/article/details/80217267

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int maxn = 200010;

int n, k;
int sa[maxn] , Rank[maxn] , height[maxn];
int p[maxn] , tmp[maxn] , cnt[maxn];
int s[maxn];
string Tmp;
bool Isequal(int x,int y,int l) {
	return (Rank[x] == Rank[y] && Rank[x + l] == Rank[y + l]);
}

void Get_SA() {
//	cout << s << " " << n << endl;
	for(int i = n; i>=0; --i)
		s[i] = s[i - 1];

	for(int i = 1; i <= n; ++i) {
		Rank[i] = s[i];
		sa[i] = i;
	}

	for(int l = 0 , pos = 0 , sig = 122; pos < n; sig = pos) {
		pos = 0;

		for(int i = n - l + 1; i <= n; ++i)
			p[++pos] = i;

		for(int i = 1; i <= n; ++i)
			if(sa[i] > l) p[ ++pos ] = sa[i] - l;

		memset( cnt , 0 ,sizeof(int) * (sig + 1));

		for(int i = 1; i <= n; ++i)
			++cnt[Rank[i]];

		for(int i = 1; i <= sig; ++i)
			cnt[i] += cnt[i - 1];

		for(int i = n; i; --i)
			sa[cnt[Rank[p[i]]] -- ] = p[i];

		pos = 0;

		for(int i = 1; i <= n; ++i)
			tmp[sa[i]] = Isequal(sa[i] , sa[i - 1] , l)?pos:++pos;

		for(int i = 1; i <= n; ++i)
			Rank[i] = tmp[i];

		l = !l? 1 : l << 1;
	}

	/*	for(int i=1; i<=n; i++) {
			cout << Rank[i] << " "  << sa[i] << endl;
		}*/

	return;
}

void CalHeight(int *r,int *sa,int n) {
//	cout << "starting" << endl;
	for(int i=1,j=1,k=1; i<=n; height[Rank[i++]]=k) {
		for(k?--k:0, j=sa[Rank[i]-1]; r[i+k] == r[j+k]; ++k);
//		cout << i <<":"<<sa[i] << ":" << Tmp.substr(sa[i]-1) << ":" << Rank[i] << ";"<< k << endl;
	}
}

int main() {
	int T;
	cin >> T;
	while(T--) {
		cin >> Tmp;

		n = Tmp.length();
		
		for(int i=0; i<n; i++){
			s[i] = Tmp[i]; 
		} s[n] = 0;
		Get_SA();

		CalHeight(s, sa, n);

		int ans = n * (n + 1) / 2;

		for(int i = 1; i <= n; i++){
//			cout << ans << "-" << height[i] << "=";
			ans -= height[i];
//			cout << ans  << endl;
		}
			
		printf("%d\n", ans);
	}

	return 0;
}