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[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
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; }
[SPOJ7258]Lexicographical Substring Search