POJ 2406 KMP/后缀数组

 

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 = 1000000 + 5;
char str[MAXN];
int Next[MAXN],len;
void getNext(){
    int i=0, k = -1;
    Next[0] = -1;
    while (i < len){
        if (k == -1 || 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[0] != '.'){
        len = strlen(str);
        getNext();
        if (Next[len] && (len % (len - Next[len])== 0)){
            printf("%d\n", len / (len-Next[len]));
        }
        else{
            printf("1\n");
        }
    }
//#ifdef LOCAL_TIME
//    cout << "[Finished in " << clock() - start << " ms]" << endl;
//#endif
    return 0;
}

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 = 10000000 + 5;
int wa[MAXN], wb[MAXN], wv[MAXN], WS[MAXN];
int c0(int *r, int a, int b)
{
    return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}
int c12(int k, int *r, int a, int b)
{
    if (k == 2) return r[a]<r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1);
    else return r[a]<r[b] || r[a] == r[b] && wv[a + 1]<wv[b + 1];
}
void sort(int *r, int *a, int *b, int n, int m)
{
    int i;
    for (i = 0; i<n; i++) wv[i] = r[a[i]];
    for (i = 0; i<m; i++) WS[i] = 0;
    for (i = 0; i<n; i++) WS[wv[i]]++;
    for (i = 1; i<m; i++) WS[i] += WS[i - 1];
    for (i = n - 1; i >= 0; 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 = 0, tb = (n + 1) / 3, tbc = 0, p;
    r[n] = r[n + 1] = 0;
    for (i = 0; i<n; i++) if (i % 3 != 0) wa[tbc++] = i;
    sort(r + 2, wa, wb, tbc, m);
    sort(r + 1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);
    for (p = 1, rn[F(wb[0])] = 0, i = 1; i<tbc; i++)
        rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;
    if (p<tbc) dc3(rn, san, tbc, p);
    else for (i = 0; i<tbc; i++) san[rn[i]] = i;
    for (i = 0; i<tbc; i++) if (san[i]<tb) wb[ta++] = san[i] * 3;
    if (n % 3 == 1) wb[ta++] = n - 1;
    sort(r, wb, wa, ta, m);
    for (i = 0; i<tbc; i++) wv[wb[i] = G(san[i])] = i;
    for (i = 0, j = 0, p = 0; i<ta && j<tbc; p++)
        sa[p] = c12(wb[j] % 3, 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 = 0;
    for (i = 1; i <= n; i++) Rank[sa[i]] = i;
    for (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 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[0] - 1, lcpval = INF; i > 0; i--){
        lcpval = min(lcpval, height[i + 1]);
        LCP[sa[i]] = lcpval;
    }
    for (int i = Rank[0] + 1, lcpval = INF; i <= len; i++){
        lcpval = min(lcpval, height[i]);
        LCP[sa[i]] = lcpval;
    }
    int ans = 1;
    for (int k = 1; k <= len; k++){
        if (len%k != 0){ 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[0] != '.'){
        len = strlen(str);
        for (int i = 0; i <= len; i++){
            if (i == len){ r[i] = 0; continue; }
            r[i] = (int)str[i];
        }
        dc3(r, sa, len + 1, 256);
        calheight(r, sa, len);
        solve();
    }
    //#ifdef LOCAL_TIME
    //    cout << "[Finished in " << clock() - start << " ms]" << endl;
    //#endif
    return 0;
}