[SPOJ7258]Lexicographical Substring Search

[SPOJ7258]Lexicographical Substring Search

试题描述

Little Daniel loves to play with strings! He always finds different ways to have fun with strings! Knowing that, his friend Kinan decided to test his skills so he gave him a string S and asked him Q questions of the form:

If all distinct substrings of string S were sorted lexicographically, which one will be the K-th smallest?

After knowing the huge number of questions Kinan will ask, Daniel figured out that he can't do this alone. Daniel, of course, knows your exceptional programming skills, so he asked you to write him a program which given S will answer Kinan's questions.

Example:

S = "aaa" (without quotes)
substrings of S are "a" , "a" , "a" , "aa" , "aa" , "aaa". The sorted list of substrings will be:
"a", "aa", "aaa".

输入

In the first line there is Kinan's string S (with length no more than 90000 characters). It contains only small letters of English alphabet. The second line contains a single integer Q (Q <= 500) , the number of questions Daniel will be asked. In the next Q lines a single integer K is given (0 < K < 2^31).

输出

Output consists of Q lines, the i-th contains a string which is the answer to the i-th asked question.

输入示例

aaa
2
2
3

输出示例

aa
aaa

题解

这题其实就是这道题的一个子问题。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 90010

char S[maxn];
int n, rank[maxn], height[maxn], sa[maxn], Ws[maxn];

bool cmp(int* a, int p1, int p2, int l) {
	if(p1 + l > n && p2 + l > n) return a[p1] == a[p2];
	if(p1 + l > n || p2 + l > n) return 0;
	return a[p1] == a[p2] && a[p1+l] == a[p2+l];
}
void ssort() {
	int *x = rank, *y = height;
	int m = 0;
	for(int i = 1; i <= n; i++) Ws[x[i] = S[i]]++, m = max(m, x[i]);
	for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
	for(int i = n; i; i--) sa[Ws[x[i]]--] = i;
	for(int j = 1, pos = 0; pos < n; j <<= 1, m = pos) {
		pos = 0;
		for(int i = n - j + 1; i <= n; i++) y[++pos] = i;
		for(int i = 1; i <= n; i++) if(sa[i] > j) y[++pos] = sa[i] - j;
		for(int i = 1; i <= m; i++) Ws[i] = 0;
		for(int i = 1; i <= n; i++) Ws[x[i]]++;
		for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
		for(int i = n; i; i--) sa[Ws[x[y[i]]]--] = y[i];
		swap(x, y); pos = 1; x[sa[1]] = 1;
		for(int i = 2; i <= n; i++) x[sa[i]] = cmp(y, sa[i], sa[i-1], j) ? pos : ++pos;
	}
	return ;
}
void calch() {
	for(int i = 1; i <= n; i++) rank[sa[i]] = i;
	for(int i = 1, j, k = 0; i <= n; height[rank[i++]] = k)
		for(k ? k-- : 0, j = sa[rank[i]-1]; S[j+k] == S[i+k]; k++);
	return ;
}

int en[maxn];

int main() {
	scanf("%s", S + 1);
	n = strlen(S + 1);
	
	ssort();
	calch();
	for(int i = 1; i <= n; i++) en[i] = n - sa[i] + 1 - height[i];
	for(int i = 1; i <= n; i++) en[i] += en[i-1];
	int q = read();
	while(q--) {
		int k = read(), p = lower_bound(en + 1, en + n + 1, k) - en;
		int l = sa[p], r = n - (en[p] - k);
		for(int i = l; i <= r; i++) putchar(S[i]); putchar('\n');
	}
	
	return 0;
}

这道题用后缀自动机也是可以滴。对于每个节点 i 我们维护一发 f(i) 表示状态 i 之后总共有多少种填法，然后我们每一位枚举填哪个字母，看后续状态数是否 ≥ k。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 180010
#define maxa 26

char S[maxn];
int n;

int rt, last, ToT, ch[maxn][maxa], par[maxn], Max[maxn], f[maxn];
void extend(int x) {
	int p = last, np = ++ToT; Max[np] = Max[p] + 1; f[np] = 1; last = np;
	while(p && !ch[p][x]) ch[p][x] = np, p = par[p];
	if(!p){ par[np] = rt; return ; }
	int q = ch[p][x];
	if(Max[q] == Max[p] + 1){ par[np] = q; return ; }
	int nq = ++ToT; Max[nq] = Max[p] + 1; f[nq] = 1;
	memcpy(ch[nq], ch[q], sizeof(ch[q]));
	par[nq] = par[q];
	par[q] = par[np] = nq;
	while(p && ch[p][x] == q) ch[p][x] = nq, p = par[p];
	return ;
}
int sa[maxn], Ws[maxn];
void build() {
	for(int i = 1; i <= ToT; i++) Ws[n-Max[i]]++;
	for(int i = 1; i <= n; i++) Ws[i] += Ws[i-1];
	for(int i = ToT; i; i--) sa[Ws[n-Max[i]]--] = i;
	for(int i = 1; i <= ToT; i++)
		for(int c = 0; c < maxa; c++) f[sa[i]] += f[ch[sa[i]][c]];
	return ;
}

int main() {
	scanf("%s", S);
	
	n = strlen(S);
	rt = last = ToT = 1;
	for(int i = 0; i < n; i++) extend(S[i] - 'a');
	build();
//	for(int i = 1; i <= ToT; i++) printf("%d%c", f[i], i < ToT ? ' ' : '\n');
	
	int q = read();
	while(q--) {
		int k = read(), p = rt;
		while(k)
			for(int c = 0; c < maxa; c++) if(ch[p][c])
				if(f[ch[p][c]] >= k) {
					putchar(c + 'a'); k--; p = ch[p][c]; break;
				}
				else k -= f[ch[p][c]];
		putchar('\n');
	}
	
	return 0;
}