[SPOJ]DISUBSTR：Distinct Substrings&[SPOJ]SUBST1：New Distinct Substrings

题面

Vjudge
Vjudge

Sol

求一个串不同子串的个数
每个子串一定是某个后缀的前缀，也就是求所有后缀不同前缀的个数
每来一个后缀(suf(i))就会有，(len-sa[i]+1)的新的前缀，又由于有(height)个重复的，那么就是(len-sa[i]+1-height)的贡献
两个就只有数组大小的区别

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1010);

int n, a[_], sa[_], rk[_], tmp[_], height[_], t[_], mx, T;
char s[_];

IL bool Cmp(RG int x, RG int y, RG int k){  return tmp[x] == tmp[y] && tmp[x + k] == tmp[y + k];  }

IL void Suffix_Sort(){
	RG int m = mx;
	for(RG int i = 0; i <= m; ++i) t[i] = 0;
	for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];
	for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
	for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
	for(RG int k = 1; k <= n; k <<= 1){
		RG int l = 0;
		for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;
		for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;
		for(RG int i = 0; i <= m; ++i) t[i] = 0;
		for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];
		for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
		for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];
		swap(rk, tmp); rk[sa[1]] = l = 1;
		for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
		if(l >= n) break;
		m = l;
	}
	for(RG int i = 1, h = 0; i <= n; ++i){
		if(h) --h;
		while(a[i + h] == a[sa[rk[i] - 1] + h]) ++h;
		height[rk[i]] = h;
	}
}

IL int Calc(){
	RG int ans = n - sa[1] + 1;
	for(RG int i = 2; i <= n; ++i) ans += n - sa[i] + 1 - height[i];
	return ans;
}

int main(RG int argc, RG char* argv[]){
	scanf("%d", &T);
	while(T--){
		scanf(" %s", s + 1);
		n = strlen(s + 1); mx = 0;
		for(RG int i = 1; i <= n; ++i) a[i] = s[i], mx = max(mx, a[i]);
		Suffix_Sort();
		printf("%d
", Calc());
	}
    return 0;
}