[SPOJ7258]Lexicographical Substring Search

試題描述

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".

輸入

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).

輸出

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

輸入示例

aaa 2 2 3

輸出示例

aa aaa

題解

這題其實就是這道題的一個子問題。

#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 90010char 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; }

這道題用后綴自動機也是可以滴。對于每個節點 i 我們維護一發 f(i) 表示狀態 i 之后總共有多少種填法，然后我們每一位枚舉填哪個字母，看后續狀態數是否 ≥ 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 26char 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; }

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轉載于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/6544662.html

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