POJ 3693 Maximum repetition substring(字尾陣列求最長重複子串)

題目大意：和spoj687類似，就是當長度相同是需要輸出一個最小的字典序的序列。

解體思路：這次需要列舉所有的從i到d = i-L/i (d = i-L%i)的位置，然後記錄保證最大值的同時，求出來字典序最小的。

Description

Input

Output

Sample Input

ccabababc
daabbccaa
#

Sample Output

Case 1: ababab
Case 2: aa

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007

#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)

using namespace std;


inline int read()
{
    char ch;
    bool flag = false;
    int a = 0;
    while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
    if(ch != '-')
    {
        a *= 10;
        a += ch - '0';
    }
    else
    {
        flag = true;
    }
    while(((ch = getchar()) >= '0') && (ch <= '9'))
    {
        a *= 10;
        a += ch - '0';
    }
    if(flag)
    {
        a = -a;
    }
    return a;
}
void write(int a)
{
    if(a < 0)
    {
        putchar('-');
        a = -a;
    }
    if(a >= 10)
    {
        write(a / 10);
    }
    putchar(a % 10 + '0');
}

const int maxn = 100050;

int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];

int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}

void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb;
    for(i = 0; i < m; i++) ws1[i] = 0;
    for(i = 0; i < n; i++) ws1[x[i] = r[i]]++;
    for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
    for(i = n-1; i >= 0; i--) sa[--ws1[x[i]]] = i;

    for(j = 1, p = 1; p < n; j <<= 1, m = p)
    {
        for(p = 0, i = n-j; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++)
            if(sa[i] >= j) y[p++] = sa[i]-j;
        for(i = 0; i < n; i++) wv[i] = x[y[i]];
        for(i = 0; i < m; i++) ws1[i] = 0;
        for(i = 0; i < n; i++) ws1[wv[i]]++;
        for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
        for(i = n-1; i >= 0; i--) sa[--ws1[wv[i]]] = y[i];
        for(swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
    }
}


int rank[maxn], height[maxn];
void calheight(int *r, int *sa, int n)
{
    int i, j, k = 0;
    for(i = 1; i <= n; i++) rank[sa[i]] = i;
    for(int i = 0; i < n; height[rank[i++]] = k)
        for(k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
    return ;
}

int dp[maxn][30];


void RMQ(int len)
{
    for(int i = 1; i <= len; i++)
        dp[i][0] = height[i];
    for(int j = 1; 1<<j <= maxn; j++)
    {
        for(int i = 1; i+(1<<j)-1 <= len; i++)
            dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
}
int lg[maxn];

int querry(int l, int r)
{
    int k = lg[r-l+1];
    return min(dp[l][k], dp[r-(1<<k)+1][k]);
}


void init()
{
    lg[0] = -1;
    for (int i = 1; i < maxn; ++i)
        lg[i] = lg[i>>1] + 1;
}
int seq[maxn];
char str[maxn];
void Del(int n, int Case)
{
    int Max = 0;
    string tmx, tmp;
    for(int i = 1; i <= n/2; i++)
    {
        for(int j = 0; i*j+i < n; j++)
        {
            int l = rank[i*j];
            int r = rank[i*j+i];
            if(l > r) swap(l, r);
            int x = querry(l+1, r);
            int y = 0;
            if(Max <= x/i + 1)
            {
                ///assign是string的一個賦值函式string.assign(const char *, int, int);
                tmp.assign(str, i*j, i*(x/i+1));
                if(x/i+1 > Max)
                {
                    Max = x/i+1;
                    tmx = tmp;
                }
                else
                {
                    int xn = tmx.size();
                    if(!xn || tmx > tmp)
                    tmx = tmp;
                }
            }

            int xp = i - x%i;
            if(xp && j)
            {
                for(int k = xp; k < i; k++)
                {
                    l = rank[i*j-k];
                    r = rank[i*j+i-k];
                    if(l > r) swap(l, r);
                    y = querry(l+1, r);
                    int yp = y/i+1;
                    if(yp < Max) continue;
                    tmp.assign(str, i*j-k, i*yp);
                    if(Max < yp)
                    {
                        tmx = tmp;
                        Max = yp;
                    }
                    else
                    {
                        int yn = tmx.size();
                        if(!yn || tmx > tmp)
                            tmx = tmp;
                    }
                }
            }
        }
    }
    printf("Case %d: ",Case);
    cout<<tmx<<endl;
}


int main()
{
    init();
    int Case = 1;
    while(scanf("%s", str) && str[0] != '#')
    {
        int n = strlen(str);
        for(int i = 0; i < n; i++) seq[i] = str[i]-'a'+1;
        seq[n] = 0;
        da(seq, sa, n+1, 27);
        calheight(seq, sa, n);
        RMQ(n);
        Del(n, Case++);
    }
    return 0;
}