uva 12012 - Detection of Extraterrestrial(KMP)

最新推荐文章于 2021-05-23 11:37:30 发布

Aurum_potestas_est

最新推荐文章于 2021-05-23 11:37:30 发布

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分类专栏： HDU Gloria KMP 文章标签： kmp

本文链接：https://blog.youkuaiyun.com/Aurum_potestas_est/article/details/78171071

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本文介绍了一种利用KMP算法寻找字符串中最长重复子串的方法，并给出了详细的实现步骤与示例代码。该算法适用于查找由某一子串连续多次重复组成的最长子串。

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Input
The first line contains T, the number of test cases (T ≤ 200). Most of the test cases are relatively
small. T lines follow, each contains a string of only small latin letters ’a’ - ’z’, whose length N is less
than 1000, without any leading or trailing whitespaces.
Output
For each test case, output a single line, which should begin with the case number counting from 1,
followed by N integers. The X-th (1-based) of them should be the maximum length of the substring
which can be written as a concatenation of X same strings. If that substring doesn’t exist, output 0
instead. See the sample for more format details.
Hint: For the second sample, the longest substring which can be written as a concatenation of 2 same
strings is “noonnoon”, “oonnoonn”, “onnoonno”, “nnoonnoo”, any of those has length 8; the longest
substring which can be written as a concatenation of 3 same strings is the string itself. As a result, the
second integer in the answer is 8 and the third integer in the answer is 12.
Sample Input
2
arisetocrat
noonnoonnoon
Sample Output
Case #1: 11 0 0 0 0 0 0 0 0 0 0
Case #2: 12 8 12 0 0 0 0 0 0 0 0 0
题意：如果有一段字符串是由某一子串连续x次重复组成的，那么这段字符串的长度记做L1，L2……Ln，第x项表示L中最大的数。（x=1,2,3……len）
思路：首先想到用kmp的next数组求循环节，但是有个问题，这道题中循环出现的子串不一定是从s[0]开始的，有可能是中间某一段有循环节，所以用一个for(i=0;i<=len-1;i++)的大循环，每次对s+i进行kmp
然后先来纪念一下自己的错误代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
char s[1005];
int nextt[1005];
int a[1005];
void getnextt(char s[])
{
    int j, k;
    int len=strlen(s);
    j = 0; k = -1; nextt[0] = -1;
    while(j < len)
        if(k == -1 || s[j] == s[k])
        {
            nextt[++j] = ++k;
        }

        else
            k = nextt[k];

}
int main()
{
    int T,Case=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        memset(a,0,sizeof(a));
        int LEN=strlen(s);
        a[1]=LEN;
        for(int i=0;i<LEN;i++)
        {

            memset(nextt,0,sizeof(nextt));
            getnextt(s+i);
            int len=strlen(s+i);
            for(int j=2;j<=len;j++)
            //从这里开始都是凭感觉写的……例如abababab求出最小循环节是2，循环次数是4，那么a[4]=2，但是怎么继续去推出a[2]=4，难到了我……
            if(j%(j-nextt[j])==0&&nextt[j]!=0)
            {
                int x=j/(j-nextt[j]);
                int y=j-nextt[j];
                while(x)
                {
                    a[x]=max(a[x],y*x);
                    x/=2;
                    y*=2;
                }

            }
        }
        printf("Case #%d:",Case++);
        for(int k=1;k<=LEN;k++)
            printf(" %d",a[k]);
        printf("\n");

    }
    return 0;
}

然后就是正确代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
char s[1005];
int nextt[1005];
int a[1005];
void getnextt(char s[])
{
    int j, k;
    int len=strlen(s);
    j = 0; k = -1; nextt[0] = -1;
    while(j < len)
        if(k == -1 || s[j] == s[k])
        {
            nextt[++j] = ++k;
        }

        else
            k = nextt[k];

}
int main()
{
    int T,Case=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        memset(a,0,sizeof(a));
        int LEN=strlen(s);
        a[1]=LEN;
        for(int i=0;i<LEN;i++)
        {

            memset(nextt,0,sizeof(nextt));
            getnextt(s+i);
            int len=strlen(s+i);
            for(int j=2;j<=len;j++)
            {
                int tmp=j;
                while(tmp)
                {
                    tmp=nextt[tmp];
                    if(j%(j-tmp)==0)
                    {
                        int x=j/(j-tmp);
                        a[x]=max(a[x],j);
                    }
                }
            }

        }
        printf("Case #%d:",Case++);
        for(int k=1;k<=LEN;k++)
            printf(" %d",a[k]);
        printf("\n");

    }
    return 0;
}