POJ 2406 KMP/后缀数组

时间：2022-08-23 15:14:24

题目链接：http://poj.org/problem?id=2406

题意:给定一个字符串，求由一个子串循环n次后可得到原串，输出n[即输出字符串的最大循环次数]

思路一：KMP求最小循环机，然后就能求出循环次数。

#define _CRT_SECURE_NO_DEPRECATE

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<string>

#include<queue>

#include<vector>

#include<time.h>

#include<cmath>

using namespace std;

typedef long long int LL;

const int MAXN =  + ;

char str[MAXN];

int Next[MAXN],len;

void getNext(){

    int i=, k = -;

    Next[] = -;

    while (i < len){

        if (k == - || str[i] == str[k]){

            ++i; ++k;

            Next[i] = k;

        }

        else{

            k = Next[k];

        }

    }

}

int main(){

//#ifdef kirito

//    freopen("in.txt","r",stdin);

//    freopen("out.txt","w",stdout);

//#endif

//    int start = clock();

    while (scanf("%s", str) && str[] != '.'){

        len = strlen(str);

        getNext();

        if (Next[len] && (len % (len - Next[len])== )){

            printf("%d\n", len / (len-Next[len]));

        }

        else{

            printf("1\n");

        }

    }

//#ifdef LOCAL_TIME

//    cout << "[Finished in " << clock() - start << " ms]" << endl;

//#endif

    return ;

}

思路二：后缀数组，直接根据09年oi论文<<后缀数组——出来字符串的有力工具>>的解法。

穷举字符串S 的长度k，然后判断是否满足。判断的时候，先看字符串L 的长度能否被k 整除，再看suffix(1)和suffix(k+1)的最长公共前缀是否等于n-k。在询问最长公共前缀的时候，suffix(1)是固定的，所以RMQ问题没有必要做所有的预处理， 只需求出height 数组中的每一个数到height[rank[1]]之间的最小值即可。整个做法的时间复杂度为O(n)。

补充：该题字符串长度比较大，达到1e7的上限，所以O(nlogn)的倍增会TLE，所以考虑O(n)的DC3。 但是还是要2500ms才能AC，对于KMP的125ms来说，该题还是KMP比较优而且代码了比较少，不过学习到求最小循环次数还可以用后缀数组来做。

#define _CRT_SECURE_NO_DEPRECATE

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<string>

#include<queue>

#include<vector>

#include<time.h>

#include<cmath>

using namespace std;

typedef long long int LL;

#define INF 0x3f3f3f3f

#define F(x) ((x)/3+((x)%3==1?0:tb))

#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)

const int MAXN =  + ;

int wa[MAXN], wb[MAXN], wv[MAXN], WS[MAXN];

int c0(int *r, int a, int b)

{

    return r[a] == r[b] && r[a + ] == r[b + ] && r[a + ] == r[b + ];

}

int c12(int k, int *r, int a, int b)

{

    if (k == ) return r[a]<r[b] || r[a] == r[b] && c12(, r, a + , b + );

    else return r[a]<r[b] || r[a] == r[b] && wv[a + ]<wv[b + ];

}

void sort(int *r, int *a, int *b, int n, int m)

{

    int i;

    for (i = ; i<n; i++) wv[i] = r[a[i]];

    for (i = ; i<m; i++) WS[i] = ;

    for (i = ; i<n; i++) WS[wv[i]]++;

    for (i = ; i<m; i++) WS[i] += WS[i - ];

    for (i = n - ; i >= ; i--) b[--WS[wv[i]]] = a[i];

    return;

}

void dc3(int *r, int *sa, int n, int m)

{

    int i, j, *rn = r + n, *san = sa + n, ta = , tb = (n + ) / , tbc = , p;

    r[n] = r[n + ] = ;

    for (i = ; i<n; i++) if (i %  != ) wa[tbc++] = i;

    sort(r + , wa, wb, tbc, m);

    sort(r + , wb, wa, tbc, m);

    sort(r, wa, wb, tbc, m);

    for (p = , rn[F(wb[])] = , i = ; i<tbc; i++)

        rn[F(wb[i])] = c0(r, wb[i - ], wb[i]) ? p -  : p++;

    if (p<tbc) dc3(rn, san, tbc, p);

    else for (i = ; i<tbc; i++) san[rn[i]] = i;

    for (i = ; i<tbc; i++) if (san[i]<tb) wb[ta++] = san[i] * ;

    if (n %  == ) wb[ta++] = n - ;

    sort(r, wb, wa, ta, m);

    for (i = ; i<tbc; i++) wv[wb[i] = G(san[i])] = i;

    for (i = , j = , p = ; i<ta && j<tbc; p++)

        sa[p] = c12(wb[j] % , r, wa[i], wb[j]) ? wa[i++] : wb[j++];

    for (; i<ta; p++) sa[p] = wa[i++];

    for (; j<tbc; p++) sa[p] = wb[j++];

    return;

}

int Rank[MAXN], height[MAXN], sa[MAXN];

void calheight(int *r, int *sa, int n){

    int i, j, k = ;

    for (i = ; i <= n; i++) Rank[sa[i]] = i;

    for (i = ; i < n; height[Rank[i++]] = k)

        for (k ? k-- : , j = sa[Rank[i] - ]; r[i + k] == r[j + k]; k++);

    return;

}

int len, r[MAXN], LCP[MAXN];

char str[MAXN];

void solve(){

    //LCP[i]:suffix(0)和suffix(i)的最长公共前缀

    for (int i = Rank[] - , lcpval = INF; i > ; i--){

        lcpval = min(lcpval, height[i + ]);

        LCP[sa[i]] = lcpval;

    }

    for (int i = Rank[] + , lcpval = INF; i <= len; i++){

        lcpval = min(lcpval, height[i]);

        LCP[sa[i]] = lcpval;

    }

    int ans = ;

    for (int k = ; k <= len; k++){

        if (len%k != ){ continue; }

        if (LCP[k] == len - k){ //第一个找到一定是最优解

            ans = len / k;

            break;

        }

    }

    printf("%d\n", ans);

}

int main(){

    //#ifdef kirito

    //    freopen("in.txt","r",stdin);

    //    freopen("out.txt","w",stdout);

    //#endif

    //    int start = clock();

    while (scanf("%s", str) && str[] != '.'){

        len = strlen(str);

        for (int i = ; i <= len; i++){

            if (i == len){ r[i] = ; continue; }

            r[i] = (int)str[i];

        }

        dc3(r, sa, len + , );

        calheight(r, sa, len);

        solve();

    }

    //#ifdef LOCAL_TIME

    //    cout << "[Finished in " << clock() - start << " ms]" << endl;

    //#endif

    return ;

}

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