HDU 1403 Longest Common Substring

Problem Description

Given two strings, you have to tell the length of the Longest Common Substring of them.

For example:
str1 = banana
str2 = cianaic

So the Longest Common Substring is "ana", and the length is 3.

 

Input

The input contains several test cases. Each test case contains two strings, each string will have at most 100000 characters. All the characters are in lower-case.

Process to the end of file.

 

Output

For each test case, you have to tell the length of the Longest Common Substring of them.

 

Sample Input

bananacianaic

 

Sample Output

3
后缀数组的应用
#include<iostream>#include<cmath>#include<map>#include<vector>#include<queue>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 200005;class suffix{private:char s[maxn];int r[maxn], w[maxn], ss[maxn], h[maxn];int sa[maxn], rk[maxn + maxn], size, bit = 256;int size1;public:bool get(){ if (scanf("%s", s + 1) != 1) return false;size1 = strlen(s + 1);scanf("%s", s + size1 + 2);s[++size1] = 1;size = strlen(s + 1);return true;}void pre(){memset(rk, 0, sizeof(rk));for (int i = 1; i <= bit; i++) w[i] = 0;for (int i = 1; i <= size; i++) w[(int)s[i]]++;for (int i = 1; i <= bit; i++) w[i] += w[i - 1];for (int i = size; i; i--) sa[w[(int)s[i]]--] = i;for (int i = 1, j = 1; i <= size; i++)rk[sa[i]] = (s[sa[i]] == s[sa[i + 1]] ? j : j++);for (int j = 1; j < size; j += j){for (int i = 1; i <= size; i++) w[i] = 0;for (int i = 1; i <= size; i++) w[rk[i + j]]++;for (int i = 1; i <= size; i++) w[i] += w[i - 1];for (int i = size; i; i--) ss[w[rk[i + j]]--] = i;for (int i = 1; i <= size; i++) w[i] = 0;for (int i = 1; i <= size; i++) w[rk[ss[i]]]++;for (int i = 1; i <= size; i++) w[i] += w[i - 1];for (int i = size; i; i--) sa[w[rk[ss[i]]]--] = ss[i];for (int i = 1, k = 1; i <= size; i++)r[sa[i]] = (rk[sa[i]] == rk[sa[i + 1]] && rk[sa[i] + j] == rk[sa[i + 1] + j]) ? k : k++;for (int i = 1; i <= size; i++) rk[i] = r[i];}for (int i = 1, k = 0, j; i <= size; h[rk[i++]] = k)for (k ? k-- : 0, j = sa[rk[i] - 1]; s[i + k] == s[j + k]; k++);}void work(){int mid = (size1 - 1) >> 1;for (int l = 0, r = size1 - 1,f; l < r;){f = 0;for (int i = 2, j; i <= size; i = j + 1){int f1 = 0, f2 = 0;for (j = i; h[j] >= mid; j++){if (sa[j] < size1) f1 = 1; else f2 = 1;if (sa[j - 1] < size1) f1 = 1; else f2 = 1;if (f1 && f2) { f = 1; goto end; }}}end:;if (f) l = mid; else r = mid - 1;mid = (l + r) >> 1;if (mid == l) mid = r;}printf("%d\n", mid);}}f;int main(){while (f.get()){f.pre();f.work();}return 0;}