G - Maximum repetition substring POJ - 3693 (后缀数组)

又是一道后缀数组的题目，和上一道题目是类似的，不过要输字符串本身，在同等重复次数下输出，字典序最小的。
我们只需要记录下所有达到最大重复次数的，重复段的长度就可以了，然后按照sa数组排序好的字典序，对每一个sa[i]，去枚举记录下来的重复长度，如果刚好达到要求直接输出就行了。

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
#define Max_N (2*50000 + 100)
int n;
int k;
int a[Max_N];
int rank1[Max_N];
int tmp[Max_N];
bool compare_sa(int i, int j)
{
    if(rank1[i] != rank1[j]) return rank1[i] < rank1[j];
    else {
        int ri = i + k <= n ? rank1[i + k] : -1;
        int rj = j + k <= n ? rank1[j + k] : -1;
        return ri < rj;
    }
}

void construct_sa(int buf[], int s, int sa[])
{
    int len = s;
    for (int i = 0; i <= len; i++) {
        sa[i] = i;
        rank1[i] = i < len ? buf[i] : -1;
    }

    for ( k = 1; k <= len; k *= 2) {
        sort(sa, sa + len +1, compare_sa);
        tmp[sa[0]] = 0;
        for (int i = 1; i <= len; i++) {
            tmp[sa[i]] = tmp[sa[i-1]] + (compare_sa(sa[i-1], sa[i]) ? 1 : 0);
        }
        for (int i = 0; i <= len; i++) {
            rank1[i] = tmp[i];
        }
    }
}

void construct_lcp(int buf[], int len, int *sa, int *lcp)
{
    int h = 0;
    lcp[0] = 0;
    for (int i = 0; i < len; i++) {
        int j = sa[rank1[i] - 1];
        if (h > 0) h--;
        for (; j + h < len && i + h < len; h++) {
            if (buf[j+h] != buf[i+h]) break;
        }
        lcp[rank1[i] - 1] = h;
    }
}
int sa[Max_N];
int rev[Max_N];
int lcp[Max_N];
char buf[Max_N];
int f[Max_N][30];
void rmq()
{
    for (int i = 1; i <= n; i++)
        f[i][0] = lcp[i];
    f[n][0] = 1000000;
     for (int j=1; (1<<j) <= n ; j++)  
        for (int i=1; i+(1<<j)-1<=n; i++)  
            f[i][j]= min(f[i][j-1],f[i+(1<<j-1)][j-1]);  
    return;  
}

int lcp1(int L, int R)
{
    if (L> R) {
        int ss = R;
        R = L;
        L = ss;
    }
    R--;
    int k1 = 0;
    while((1<<(k1+1)) <= R-L+1) k1++;
    return min(f[L][k1], f[R-(1<<k1)+1][k1]);
}
vector<int> v;
int main()
{
    int T;
    T = 0;
    while (true) {
        scanf("%s", buf);
        if (buf[0] == '#') return 0;
        n = strlen(buf);
        for (int i = 0; i < n; i++)
            a[i] = int(buf[i]);
        a[n] = 0;
        construct_sa(a, n, sa);
        construct_lcp(a, n, sa, lcp);
        rmq();
        /*int l,r;
        cin >> l >> r;
        cout << lcp1(l, r) << endl;*/
        int max = 0;
        v.clear();
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j + i < n; j += i) {
                int k1 = lcp1(rank1[j], rank1[j+i]);
                int r = k1 / i + 1;
                int t = j - (i - k1%i);
                if (t >= 0)
                    if (lcp1(rank1[t], rank1[t+i]) >= i) {
                        r++;
                    }
                if (r > max) {
                    max = r;
                    v.clear();
                    v.push_back(i);

                }   
                if (r == max) {
                    v.push_back(i);
                }

                }
            }
        int beg = 0;
        int len = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < v.size(); j++) {
                if (sa[i] + v[j] <= n) {
                    if (lcp1(i, rank1[sa[i]+v[j]]) >= (max-1) * v[j]) {
                        beg = sa[i];
                        len = v[j] * max;
                        i = n+1;
                        j = v.size() + 1;
                    }
                }
            }
        }
        printf("Case %d: ", ++T);
        for (int i = beg; i < beg + len; i++)
            printf("%c", buf[i]);
        cout << endl;
    }
    return 0;
}