POJ - 3693 Maximum repetition substring (后缀数组)

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>

#define fuck(x) cerr<<#x<<" = "<<x<<endl;
#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define ls (t<<1)
#define rs ((t<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);

char s[maxn];
int len, Rank[maxn], sa[maxn], tlen, tmp[maxn];
bool compare_sa(int i, int j) {
    if (Rank[i] != Rank[j]) { return Rank[i] < Rank[j]; }
    //如果以i开始,长度为k的字符串的长度,已经超出了字符串尾,那么就赋值为-1
    //这是因为,在前面所有数据相同的情况下,字符串短的字典序小.
    int ri = i + tlen <= len ? Rank[i + tlen] : -inf;
    int rj = j + tlen <= len ? Rank[j + tlen] : -inf;
    return ri < rj;
}
void construct_sa() {
    //初始的RANK为字符的ASCII码
    for (int i = 0; i <= len; i++) {
        sa[i] = i;
        Rank[i] = i < len ? s[i] : -inf;
    }
    for (tlen = 1; tlen <= len; tlen *= 2) {
        sort(sa, sa + len + 1, compare_sa);
        tmp[sa[0]] = 0;
        //全新版本的RANK,tmp用来计算新的rank
        //将字典序最小的后缀rank计为0
        //sa之中表示的后缀都是有序的,所以将下一个后缀与前一个后缀比较,如果大于前一个后缀,rank就比前一个加一.
        //否则就和前一个相等.
        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++) {
            Rank[i] = tmp[i];

        }
    }
}
int height[maxn];
void construct_lcp() {
//    for(int i=0;i<=n;i++){Rank[sa[i]]=i;}
    int h = 0;
    height[0] = 0;
    for (int i = 0; i < len; i++) {//i为后缀数组起始位置
        int j = sa[Rank[i] - 1];//获取当前后缀的前一个后缀(排序后)
        if (h > 0)h--;
        for (; j + h < len && i + h < len; h++) {
            if (s[j + h] != s[i + h])break;
        }
        height[Rank[i]] = h;
    }
}
int st[maxn][20];
void rmq_init() {
    for (int i = 1; i <= len; i++) {
        st[i][0] = height[i];
    }
    int l = 2;
    for (int i = 1; l <= len; i++) {
        for (int j = 1; j + l / 2 <= len; j++) {
            st[j][i] = min(st[j][i - 1], st[j + l / 2][i - 1]);
        }
        l <<= 1;
    }
}
int ask_min(int i, int j) {
    int k = int(log(j - i + 1.0) / log(2.0));
    return min(st[i][k], st[j - (1 << k) + 1][k]);
}
int lcp(int a, int b) {//此处参数是,原字符串下标
    a = Rank[a], b = Rank[b];
    if (a > b)
        swap(a, b);
    return ask_min(a + 1, b);
}
vector<int>ansl;
int main() {
    int cases=0;
    while (scanf("%s",s)!=EOF){
        ansl.clear();
        len = strlen(s);
        cases++;
        if(len==1&&s[0]==‘#‘){ break;}
        construct_sa();
        construct_lcp();
        rmq_init();
        int ansx,ans;
        ansx=ans=0;
        for(int i=1;i<=len;i++){
            bool flag = true;
            for(int j=0;j+i<len;j+=i){
                ans=lcp(j,j+i);
                int k=j-(i-ans%i);
                ans=ans/i+1;
                if(k>=0&&lcp(k,k+i)>=i){ans++;}

                if(ans==ansx){
                    if(flag){
                        ansl.push_back(i);
                        flag = false;
                    }
                }else if(ans>ansx){
                    ansx=ans;
                    ansl.clear();
                    ansl.push_back(i);
                    flag=false;
                }
            }
        }

        int siz = ansl.size();
        bool flag = false;
        for(int i=1;i<=len;i++){
            for(int j=0;j<siz;j++){
                int l = ansl[j];
                if(sa[i]+l<len&&lcp(sa[i],sa[i]+l)>=(ansx-1)*l){//核心代码,如果sa[i]和sa[i]+l的公共前缀中包含了ansx-1个l,
                    // 说明sa[i]的前缀中已经包含了ansx个l
                    ans = sa[i];
                    s[ans+ansx*l]=0;
                    flag=true;
                }
            }
            if(flag){ break;}
        }
        printf("Case %d: %s\n",cases,s+ans);
    }

    return 0;
}