FZU 1901 Period II (KMP)

最新推荐文章于 2019-11-18 21:28:29 发布

原创最新推荐文章于 2019-11-18 21:28:29 发布 · 368 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#fzu #kmp

KMP 专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一种用于找出字符串所有周期性前缀的算法，并提供了详细的实现步骤及C++代码示例。该算法通过构建next数组来快速定位周期性前缀。

Problem 1901 Period II

Accept: 518 Submit: 1270
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

For each prefix with length P of a given string S,if

S[i]=S[i+P] for i in [0..SIZE(S)-p-1],

then the prefix is a “period” of S. We want to all the periodic prefixs.

Input

Input contains multiple cases.

The first line contains an integer T representing the number of cases. Then following T cases.

Each test case contains a string S (1 <= SIZE(S) <= 1000000),represents the title.S consists of lowercase ,uppercase letter.

Output

For each test case, first output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of periodic prefixs.Then output the lengths of the periodic prefixs in ascending order.

Sample Input

4oooacmacmacmacmacmafzufzufzufstostootssto

Sample Output

Case #1: 31 2 3Case #2: 63 6 9 12 15 16Case #3: 43 6 9 10Case #4: 29 12

Source

FOJ有奖月赛-2010年05月

思路：对于任意i属于[0,size(s)-p-1]要满足s[i]=s[i+p]，所以必有s[size(s)-p-1]=s[size(s)-1]，即s[size(s)-p-1]就是next数组中存的匹配的那个位置。所以可以循环直到不再匹配为止。用s[size(s)-1]减去每一次循环中得到的位置就是所求的p长度。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX_N 1000005
using namespace std;
char p[MAX_N];
int ne[MAX_N],ans[MAX_N];;
void makeNext(const char p[],int ne[])
{
    int len=strlen(p);
    ne[0]=0;
    for(int i=1,k=0;i<len;i++)
    {
        while(k>0&&p[i]!=p[k]) k=ne[k-1];
        if(p[i]==p[k]) k++;
        ne[i]=k;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int k=1;k<=t;k++)
    {
        scanf("%s",p);
        memset(ne,0,sizeof(ne));
        makeNext(p,ne);
        int len=strlen(p),f=ne[len-1],cnt=0;
        while(f>0)
        {
            ans[++cnt]=len-f;
            f=ne[f-1];
        }
        ans[++cnt]=len;
        printf("Case #%d: %d\n",k,cnt);
        for(int i=1;i<=cnt;i++)
        {
            if(i==cnt)
            printf("%d\n",ans[i]);
            else printf("%d ",ans[i]);
        }
    }
    return 0;
}