poj 3415 Common Substrings
阿新 • • 發佈:2018-05-20
top != accepted lang dep ons pri tro vector Common Substrings
T(i, k)=TiTi+1...Ti+k-1, 1≤i≤i+k-1≤|T|.
S = {(i, j , k) | k≥K, A(i, k)=B(j, k)}.
|, |B| ≤ 105
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 12585 | Accepted: 4228 |
Description
A substring of a string T is defined as:
Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):
You are to give the value of |S| for specific A, B and K.
Input
The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.
1 ≤ |A
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.
Output
For each case, output an integer |S|.
Sample Input
2 aababaa abaabaa 1 xx xx 0
Sample Output
22 5
題意:求兩個字符串的長度大於k的子串的數量
思路:其實就是求兩個字符串當中的任意兩個後綴的相同前綴的數量,設lcp是任意兩個後綴的相同前綴的最大長度,那麽這兩個後綴的長度大於K的相同前綴數量為lcp-K+1.
直接枚舉兩個字符串的所有後綴並累加他們的前綴數量復雜度在O(n^2)行不通。
可以利用單調棧。首先把兩個字符串s1,s2進行合並,中間可以加個不同的字符(譬如‘$‘)來區別,即s=s1+‘$‘+s2 ,求s的後綴數組和高度數組。
首先任意兩個後綴,記它們在後綴數組中位置分別為i,j,則它們的高度lcp可以表示為min(lcp[i],lcp[i+1],...,lcp[j-1]),既然如此,可以用單調棧來維護lcp
對於s2的每一個後綴B,考慮所有字典序在B前面的s1的後綴Ai,計算所有Ai與B的相同前綴的數量和,可以用單調棧優化。對於s1中的每個後綴A,計算Bi與A的相同前綴數量和與之前是類似的。
在高度數組當中把高度大於等於K的連續的序列分成一塊,一塊一塊的用單調棧考慮,具體見代碼:
AC代碼:
#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<algorithm> #include<vector> #include<cstring> #include<string> #include<cmath> using namespace std; const int INF = 0x3f3f3f3f; const int N_MAX = 100000 + 20; typedef long long ll; int n, k; int Rank[N_MAX*2]; int tmp[N_MAX*2]; int sa[N_MAX * 2]; int lcp[N_MAX*2]; bool compare_sa(const int& i,const int& j) { if (Rank[i] != Rank[j])return Rank[i] < Rank[j]; else { int ri = i + k <= n ? Rank[i + k] : -1; int rj = j + k <= n ? Rank[j + k] : -1; return ri < rj; } } void construct_sa(const string& S,int *sa) { n = S.size(); for (int i = 0; i <= n;i++) { sa[i] = i; Rank[i] = i < n ? S[i] : -1; } for (k = 1; k <= n;k*=2) { sort(sa,sa+n+1,compare_sa); tmp[sa[0]] = 0; for (int i = 1; i <= n;i++) { tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0); } for (int i = 0; i <= n;i++) { Rank[i] = tmp[i]; } } } void construct_lcp(const string& S,int *sa,int *lcp){ memset(lcp,0,sizeof(lcp)); int n = S.length(); for (int i = 0; i <= n; i++)Rank[sa[i]] = i; int h = 0; lcp[0] = 0; for (int i = 0; i < n; i++) { int j = sa[Rank[i] - 1]; if (h > 0)h--; for (; j + h < n&&i + h < n; h++) { if (S[j + h] != S[i + h])break; } lcp[Rank[i] - 1] = h; } } int K; string s1, s2, s; ll top, accumu; int stack[N_MAX * 2][2];//1存放人數,0存放lcp ll find_num(int sz1,bool is_s1) { ll res = 0; top = accumu = 0; for (int i = 0; i < s.size(); i++) { if (lcp[i] < K) { top = 0; accumu = 0; } else { int size = 0;//統計高度為lcp[i]的人數 if ((is_s1&&sa[i] < sz1) || (!is_s1&&sa[i] > sz1)) {//如果是s1中的後綴 size++; accumu += lcp[i] - K + 1; } while (top>0&&lcp[i]<=stack[top-1][0]) {//前面的lcp高度比較高,則要削減高度直到和lcp[i]一樣,這樣之前的那些人的高度也變成lcp[i]了 top--; accumu -= stack[top][1] * (stack[top][0] - lcp[i]); size += stack[top][1]; } if (size) { stack[top][0] = lcp[i]; stack[top][1] = size; top++;//!!! } if ((is_s1&&sa[i+1] > sz1) || (!is_s1&&sa[i+1] < sz1)) {//sa[i+1]是s2中的後綴!!! res += accumu; } } } return res; } int main() { while (scanf("%d",&K)&&K) { cin >> s1 >> s2; int sz1 = s1.size(); int sz2 = s2.size(); s = s1 + ‘$‘ + s2; construct_sa(s,sa); construct_lcp(s,sa,lcp); printf("%lld\n",find_num(sz1,1)+find_num(sz1,0)); } return 0; }
poj 3415 Common Substrings