POJ 3261 可重叠的 k 次最长重复子串【后缀数组】

这也是一道例题

给定一个字符串，求至少出现 k 次的最长重复子串，这 k 个子串可以重叠。
算法分析：
这题的做法和上一题差不多，也是先二分答案，然后将后缀分成若干组。不
同的是，这里要判断的是有没有一个组的后缀个数不小于 k。如果有，那么存在
k 个相同的子串满足条件，否则不存在。这个做法的时间复杂度为 O(nlogn)。

Source Code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int N = 1              ;
const int M = 200000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;

int a[M], sa[M], h[M], rank[M];

void radix(int *str, int *a, int *b, int n, int m){
    static int count[M];
    int i;
    memset(count, 0, sizeof(count));
    for(i = 0; i < n; ++i)  ++count[str[a[i]]];
    for(i = 1; i <= m; ++i) count[i] += count[i - 1];
    for(i = n - 1; i >= 0; --i) b[--count[str[a[i]]]] = a[i];
}

void suffix_array(int *str, int *sa, int n, int m){
    static int a[M], b[M];
    int i, j;
    for(i = 0; i < n; ++i)  rank[i] = i;
    radix(str, rank, sa, n, m);

    rank[sa[0]] = 0;
    for(i = 1; i < n; ++i)  rank[sa[i]] = rank[sa[i - 1]] + (str[sa[i]] != str[sa[i - 1]]);
    for(i = 0; 1<<i < n; ++i){
        for(j = 0; j < n; ++j){
            a[j] = rank[j] + 1;
            b[j] = j + (1 << i) >= n ? 0 : rank[j + (1 << i)] + 1;
            sa[j] = j;
        }
        radix(b, sa, rank, n, n);
        radix(a, rank, sa, n, n);
        rank[sa[0]] = 0;
        for(j = 1; j < n; ++j){
            rank[sa[j]] = rank[sa[j - 1]] + (a[sa[j - 1]] != a[sa[j]] || b[sa[j - 1]] != b[sa[j]]);
        }
    }
}

void calc_height(int *str, int *sa, int *h, int n){
    int i, k = 0;
    h[0] = 0;
    for(i = 0; i < n; ++i){
        k = k == 0 ? 0 : k - 1;
        if(rank[i] != 0){
            while(str[i + k] == str[sa[rank[i] - 1] + k]){
                ++k;
            }
        }
        h[rank[i]] = k;
    }
}

void solve_duplicate_substr(int n){
    int i, j, pos, ans = 0;
    for(i = 0; i < n; ++i){
        if(h[rank[i]] > ans){
            ans = h[rank[i]];
            pos = i;
        }
    }
    for(i = pos; i < pos + ans; ++i){
        printf("%c", a[i]);
    }
    printf("\n");
}

void slove_update_duplicate_substr(int n, int k){
    int i, j;
    int low = 1, high = n;
    int ans = 0, pos1 = 0, pos2 = 0;
    while(low <= high){
        int mid = (low + high) / 2;
        bool flag = false;
        for(i = 0; i < n; ++i){
            if(h[i] >= mid){
                ++ans;
                if(ans >= k)    flag = true;
            }
            else    ans = 1;
        }
        if(flag)    low = mid + 1;
        else    high = mid - 1;
    }
    cout << high << endl;
}

int main(){
    int i, j, t, n, m, k;
    while(cin >> n >> k){
        for(i = 0; i < n; ++i)  cin >> a[i];
        suffix_array(a, sa, n, 256);
        calc_height(a, sa, h, n);
        slove_update_duplicate_substr(n, k);
    }
    return 0;
}