ZOJ 3199 Longest Repeated Substring

Longest Repeated Substring

Time Limit: 5000ms

Memory Limit: 32768KB

This problem will be judged on ZJU. Original ID: 3199
64-bit integer IO format: %lld      Java class name: Main

 

Write a program that takes a string and returns length of the longest repeated substring. A repeated substring is a sequence of characters that is immediately followed by itself.

For example, given "Mississippi", the longest repeated substring is "iss" or "ssi" (not "issi").
Given "Massachusetts", the longest repeated substring would be either "s" or "t".
Given "Maine", the longest repeated substring is "" (the empty string).

Input

The first line of the input contains a single integer T , the number of test cases.

Each of the following T lines, is exactly one string of lowercase charactors.

The length of each string is at most 50000 characters.

Output

For each test case, print the length of the Longest Repeated Substring.

Sample Input

2
aaabcabc
ab

Sample Output

3
0

 

Source

ZOJ Monthly, May 2009

Author

PENG, Peng

 

解题：后缀数组，这个题貌似后缀自动机不好搞

 

直接爆

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
char s[maxn];
int sa[maxn],t[maxn],t2[maxn];
int height[maxn],rk[maxn],c[maxn],n;
void build_sa(int m) {
    int i,*x = t,*y = t2;
    for(i = 0; i < m; ++i) c[i] = 0;
    for(i = 0; i < n; ++i) c[x[i] = s[i]]++;
    for(i = 1; i < m; ++i) c[i] += c[i-1];
    for(i = n-1; i >= 0; --i) sa[--c[x[i]]] = i;

    for(int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for(i = n-k; i < n; ++i) y[p++] = i;
        for(i = 0; i < n; ++i)
            if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; ++i) c[i] = 0;
        for(i = 0; i < n; ++i) c[x[y[i]]]++;
        for(i = 1; i < m; ++i) c[i] += c[i-1];
        for(i = n-1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        x[sa[0]] = 0;
        for(p = i = 1; i < n; ++i)
            if(y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k])
                x[sa[i]] = p-1;
            else x[sa[i]] = p++;
        if(p >= n) break;
        m = p;
    }
}
void getHeight() {
    int i,j,k = 0;
    for(i = 0; i < n; ++i) rk[sa[i]] = i;
    for(i = 0; i < n; ++i) {
        if(k) --k;
        j = sa[rk[i]-1];
        while(i + k < n && j + k < n && s[i+k] == s[j+k]) ++k;
        height[rk[i]] = k;
    }
}
int main(){
    int kase;
    scanf("%d",&kase);
    while(kase--){
        scanf("%s",s);
        n = strlen(s) + 1;
        build_sa(128);
        getHeight();
        int ret = 0;
        for(int i = 2; i < n; ++i){
            int tmp = height[i];
            if(abs(sa[i-1] - sa[i]) == tmp && tmp > ret) ret = tmp;
            for(int j = i + 1; j < n; ++j){
                tmp = min(tmp,height[j]);
                if(tmp <= ret) break;
                if(abs(sa[i-1] - sa[i]) == tmp && tmp > ret) ret = tmp;
            }
        }
        printf("%d\n",ret);
    }
    return 0;
}