Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Mr. Kim is a professional programmer. Recently he wants to design a new editor which has as many functions as possible. Most editors support a simple search function that finds one occurrence (or all occurrences successively) of a query pattern string in the text.

He observed that the search function in commercial editors does nothing if no query pattern is given. His idea of a new search function regards each substring of the given text as a query pattern string itself and his new function finds another occurrence in the text. The problem is that there can be occurrences of many substrings in the text. So, Mr. Kim decides that the new function finds only occurrences of the longest substring in the text in order to remedy the problem. A formal definition of the search function is as follows:

Given a text string S , find the longest substring in text string S such that the substring appears at least twice. The two occurrences are allowed to overlap.

Input 

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. For each test case, a text string S is given in one line. For every string, the length is less than or equal to 5,000 and the alphabet is the set of lowercase English characters.

Output 

Your program is to write to standard output. Print exactly one line for each test case. Print the length of the longest substring in text string S such that the substring appears at least twice.

Sample Input 

3 
abcdefghikjlmn 
abcabcabc 
abcdabcabb

Sample Output 

0 
6 
3

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct SuffixArray {
    int sa[5005], h[5005], n;
    int w[5005], ta[5005], tb[5005];
    char str[5005];
    void sort(int *x, int *y, int m) {
        static int i;
        for(i = 0; i < m; i++)
            w[i] = 0;
        for(i = 0; i < n; i++)
            w[x[y[i]]]++;
        for(i = 1; i < m; i++)
            w[i] += w[i-1];
        for(i = n-1; i >= 0; i--)
            sa[--w[x[y[i]]]] = y[i];
    }
    bool cmp(int *r, int a, int b, int l) {
        if(r[a] == r[b]) {
            if(a+l >= n || b+l >= n)
                return false;
            return r[a+l] == r[b+l];
        }
        return false;
    }
    void build_h() {
        int i, j, k;
        for(i = 0; i < n; i++)  ta[sa[i]] = i;
        for(i = 0; i < n; i++) {
            if(ta[i] == 0) {
                h[ta[i]] = 0;
                continue;
            }
            if(i == 0 || h[ta[i-1]] <= 1)
                k = 0;
            else
                k = h[ta[i-1]]-1;
            while(str[sa[ta[i]-1]+k] == str[sa[ta[i]]+k])
                k++;
            h[ta[i]] = k;
        }
    }
    void build() {// x: rank, y: second key
        int i, k, m = 128, p;
        int *x = ta, *y = tb, *z;
        n = strlen(str);
        x[n] = 0;
        for(i = 0; i < n; i++)
            x[i] = str[i], y[i] = i;
        sort(x, y, m);
        for(k = 1, p = 1; p < n; k *= 2, m = p) {
            for(p = 0, i = n-k; i < n; i++)
                y[p++] = i;
            for(i = 0; i < n; i++) {
                if(sa[i] >= k) {
                    y[p++] = sa[i]-k;
                }
            }
            sort(x, y, m);
            z = x, x = y, y = z;
            for(i = 1, p = 1, x[sa[0]] = 0; i < n; i++)
                x[sa[i]] = cmp(y, sa[i-1], sa[i], k) ? p-1 : p++;
        }
    }
};
int main() {
    SuffixArray in;
    int t;
    scanf("%d", &t);
    getchar();
    while(t--) {
        gets(in.str);
        in.build();
        in.build_h();
        int ans = 0, anst = 0;
        for(int i = 0; i < in.n; i++)
            ans = ans > in.h[i] ? ans : in.h[i];
        printf("%d\n", ans);
    }
    return 0;
}