HDU4622 Reincarnation

vjudge传送


给一个字符串，多次询问([l,r])中不同子串的个数。((1 leqslant |S| leqslant 2000))


因为串长才2000，所以可以考虑(O(n^2))的做法。
令(ans[l][r])表示区间([l,r])中不同子串的个数，考虑怎么递推：
我们建立(n)个SAM，第(i)个SAM表示([S_i,S_n])。往第(i)个SAM插入位置为(j)的字符后，最后的节点为(u)，那么(ans[i][j]=ans[i][j-1]+len[fa] - len[u]).
这样做的好处在于，SAM能很容易的表示子串的尾位置(S_r)，如果我们限定了(S_l)，就能确定每一个子串了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const int maxn = 2e3 + 5;
const int maxs = 27;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}

char s[maxn];

int ans[maxn][maxn];
struct Sam
{
	int tra[maxn << 1][maxs], link[maxn << 1], len[maxn << 1], cnt, las;
	In void init() 
	{
		link[cnt = las = 0] = -1; Mem(tra[0], 0);
	}	
	In void insert(int c)
	{
		int now = ++cnt, p = las; Mem(tra[now], 0);
		len[now] = len[p] + 1; 
		while(~p && !tra[p][c]) tra[p][c] = now, p = link[p];
		if(p == -1) link[now] = 0;
		else
		{
			int q = tra[p][c];
			if(len[q] == len[p] + 1) link[now] = q;
			else
			{
				int clo = ++cnt;
				memcpy(tra[clo], tra[q], sizeof(tra[q]));
				len[clo] = len[p] + 1;
				link[clo] = link[q]; link[q] = link[now] = clo;
				while(~p && tra[p][c] == q) tra[p][c] = clo, p = link[p];
			}
		}
		las = now;
	}
}S;

int main()
{
	int T = read();
	while(T--)
	{
		Mem(ans, 0);
		scanf("%s", s);
		int n = strlen(s);
		for(int i = 0; i < n; ++i)
		{
			S.init();
			for(int j = i; j < n; ++j)
			{
				S.insert(s[j] - 'a');
				ans[i + 1][j + 1] = ans[i + 1][j] + S.len[S.las] - S.len[S.link[S.las]];	
			} 
		}
		int m = read();
		for(int i = 1; i <= m; ++i)
		{
			int L = read(), R = read();
			write(ans[L][R]), enter;
		}
	}
	return 0;
}