Repeats SPOJ - REPEATS

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概述

A string s is called an (k,l)-repeat if s is obtained by concatenating k>=1 times some seed string t with length l>=1. For example, the string

s = abaabaabaaba

is a (4,3)-repeat with t = aba as its seed string. That is, the seed string t is 3 characters long, and the whole string s is obtained by repeating t 4 times.

Write a program for the following task: Your program is given a long string u consisting of characters ‘a’ and/or ‘b’ as input. Your program must find some (k,l)-repeat that occurs as substring within u with k as large as possible. For example, the input string

u = babbabaabaabaabab

contains the underlined (4,3)-repeat s starting at position 5. Since u contains no other contiguous substring with more than 4 repeats, your program must output the maximum k.

Input
In the first line of the input contains H- the number of test cases (H <= 20). H test cases follow. First line of each test cases is n - length of the input string (n <= 50000), The next n lines contain the input string, one character (either ‘a’ or ‘b’) per line, in order.

Output
For each test cases, you should write exactly one interger k in a line - the repeat count that is maximized.

Example
Input:
1
17
b
a
b
b
a
b
a
a
b
a
a
b
a
a
b
a
b

Output:
4
since a (4, 3)-repeat is found starting at the 5th character of the input string.
SPOJ REPEATS Repeats （后缀数组 + RMQ：子串的最大循环节）题解
题意：
给定一个串s，s必有一个最大循环节的连续子串ss，问最大循环次数是多少

思路：
我们可以知道，如果一个长度为L的子串连续出现了两次及以上，那么必然会存在s[0]、s[L]、s[2L]⋯s[L∗k]中至少有两个连续的位置是相同的，然后看字母s[L∗i]和s[L∗(i+1)]往前往后最多能匹配多远，记住总长度len，那么最大循环次数为(len/L)+1。

参考：
SPOJ 687. Repeats(后缀数组)

代码：

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 50000 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 11;
const int MOD = 1e9 + 7;
using namespace std;
 
//下标从0开始
int str[maxn];  //str[n]赋值一个最小值0,其他大于0
int t1[maxn], t2[maxn], c[maxn];
int sa[maxn];   //排名为i的后缀下标
int rk[maxn];   //后缀下标为i的排名
int height[maxn];   //sa[i]与sa[i - 1]的LCP
int mm[maxn];
int dp[maxn][30];
bool cmp(int *r, int a, int b, int l){
    return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *str, int n, int m){
    n++;
    int i, j, p, *x = t1, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = str[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i - 1];
    for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(j = 1; j <= n; j <<= 1){
        p = 0;
        for(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 < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for(i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j)? p - 1 : p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    n--;
    for(i = 0; i <= n; i++) rk[sa[i]] = i;
    for(i = 0; i < n; i++){
        if(k) k--;
        j = sa[rk[i] - 1];
        while(str[i + k] == str[j + k]) k++;
        height[rk[i]] = k;
    }
}
void initRMQ(int n){
    mm[0] = -1;
    for(int i = 1; i <= n; i++){
        dp[i][0] = height[i];
        mm[i] = ((i & (i - 1)) == 0)? mm[i - 1] + 1 : mm[i - 1];
    }
    for(int j = 1; j <= mm[n]; j++)
        for(int i = 1; i + (1 << j) - 1 <= n; i++)
            dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
int RMQ(int L, int R){
    int k = mm[R - L + 1];
    return min(dp[L][k], dp[R - (1 << k) + 1][k]);
}
int LCP(int i, int j){   //求后缀i和j的LCP最长公共前缀
    int L = rk[i], R = rk[j];
    if(L > R) swap(L, R);
    L++;
    return RMQ(L, R);
}
 
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            char ss[2];
            scanf("%s", ss);
            str[i] = ss[0] - 'a' + 1;
        }
        str[n] = 0;
        da(str, n, 3);
        initRMQ(n);
        int ans = 1;
        for(int i = 1; i < n; i++){
            for(int j = 0; j + i < n; j += i){
                int len = LCP(j, j + i);
                int times = len / i + 1;
                int pos = j - (i - len % i);
                if(pos >= 0){
                    len = LCP(pos, pos + i);
                    times = max(times, len / i + 1);
                }
                ans = max(ans, times);
            }
        }
        printf("%dn", ans);
    }
    return 0;
}