KMP算法

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最新推荐文章于 2024-11-16 09:45:51 发布

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Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29792		Accepted: 11998

Description

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a textT, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

求模式串在主串中出现的次数

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 1000010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}


char s[10010],t[MAXN];
int slen,tlen;
int next[10010];//s为模式串  t为主串
void getNext()
{
    int j=0,k=-1;
    next[0]=-1;
    while(j<slen)
    {
        if(k==-1||s[j]==s[k])
            next[++j]=++k;
        else
            k=next[k];
    }
}

int KMP_Count()
{
    int ans=0;
    int i,j=0;
    if(slen==1&&tlen==1)
    {
        if(s[0]==t[0]) return 1;
        else return 0;
    }
    getNext();
    for(i=0;i<tlen;i++)
    {
        while(j>0&&t[i]!=s[j])
            j=next[j];
        if(s[j]==t[i]) j++;
        if(j==slen)
        {
            ans++;
            j=next[j];
        }
    }
    return ans;
}

int main()
{
//    fread;
    int tc;
    scanf("%d",&tc);
    while(tc--)
    {
        scanf("%s%s",s,t);
        slen=strlen(s);
        tlen=strlen(t);
        int ans=KMP_Count();
        printf("%d\n",ans);
    }
    return 0;
}

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15948 Accepted Submission(s): 7030

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

求模式串在主串中第一次出现的位置（从1开始算）

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

int n,m;
int a[1000010],b[10010];
int NEXT[10010];
void getNEXT()
{
    int j=0,k=-1;
    NEXT[0]=-1;
    while(j<m)
    {
        if(k==-1||b[j]==b[k])
            NEXT[++j]=++k;
        else
            k=NEXT[k];
    }
}

//返回首次出现的位置
int KMP_Index()
{
    int i=0,j=0;
    getNEXT();
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
        }
        else
            j=NEXT[j];
    }
    if(j==m) return i-m+1;
    else return -1;
}

int main()
{
//    fread;
    int tc;
    scanf("%d",&tc);
    while(tc--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&b[i]);
        int ans=KMP_Index();
        printf("%d\n",ans);
    }
    return 0;
}

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38422		Accepted: 15947

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

给定一个字符串求这个字符串最多由几个相同的字符串组成

next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}


char ch[1000010];
int next[1000010];
int len;

void getNext()
{
    int j=0,k=-1;
    next[0]=-1;
    while(j<len)
    {
        if(k==-1||ch[j]==ch[k])
            next[++j]=++k;
        else
            k=next[k];
    }
}
int main()
{
//    fread;
    while(scanf("%s",ch)!=EOF)
    {
        if(strcmp(ch,".")==0) break;
        len=strlen(ch);
        getNext();
        if(len%(len-next[len])==0&&len/(len-next[len])>1)
            printf("%d\n",len/(len-next[len]));
        else
            puts("1");
    }
    return 0;
}