KMP code implementation (no explanation and ideas)

Code:

#include <iostream>
#include <cstring>

using namespace std;

const int N = 1000100;

int n, m;
// n main string length, m pattern string length
char a[N], b[N];
// a main string, b pattern string, subscripts start at zero
int nextt[N];

void get_nextt()
{
    int i = 0, j;
    j = nextt[0] = -1;
    while(i < m)
    {
        if(-1 == j || b[i] == b[j] )
            nextt[++i] = ++j;
        else
        j = nextt[j];
    }
}

void get_nextval() {
    int i = 0, j;
    j = nextt[0] = -1;
    while(i < m) {
         if(-1 == j || b[i] == b[j]) {
            ++i, ++j;
            nextt[i] = (b[i] != b[j]) ? j : nextt[j];
         }
         else {
            j = nextt[j];
         }
    }
}

int kmp_index(int pos) {
    // Find the position of the pattern string after the pos position character of the main string
    int i = pos, j = 0;
    while(i < n && j < m) {
        if(-1 == j || a[i] == b[j]) {
            ++i, ++j;
        }
        else {
            j = nextt[j];
        }
    }
    if(j >= m) return i-m;
    else return -1;
}

int kmp_count(int pos) {
    // Find how many times the pattern string appears in the main string
    int i = pos, j = 0;
    int years = 0;
    while(i < n) {
        if(-1 == j || a[i] == b[j]) {
            ++i, ++j;
            if(j >= m) {
                ++ years;
                j = nextt[j];
            }
        }
        else
            j = nextt[j];
    }
    return ans;
}
intmain()
{
     cin >> a >> b;
     get_nextt();
     get_nextval();
     int Index = kmp_index( 0 );
     int Count = kmp_count( 0 );
     return 0;
}

Check it out with two questions:

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35581    Accepted Submission(s): 14734

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

 Code:

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 1000100;

int n, m;
// n main string length, m pattern string length
int a[N], b[N];
// a main string, b pattern string, subscripts start at zero
int nextt[N];

void get_nextt()
{
    int i = 0, j;
    j = nextt[0] = -1;
    while(i < m)
    {
        if(-1 == j || b[i] == b[j] )
            nextt[++i] = ++j;
        else
        j = nextt[j];
    }
}

void get_nextval() {
    int i = 0, j;
    j = nextt[0] = -1;
    while(i < m) {
         if(-1 == j || b[i] == b[j]) {
            ++i, ++j;
            nextt[i] = (b[i] != b[j]) ? j : nextt[j];
         }
         else {
            j = nextt[j];
         }
    }
}

int kmp_index(int pos) {
    // Find the position of the pattern string after the pos position character of the main string
    int i = pos, j = 0;
    while(i < n && j < m) {
        if(-1 == j || a[i] == b[j]) {
            ++i, ++j;
        }
        else {
            j = nextt[j];
        }
    }
    if(j >= m) return i-m;
    else return -1;
}

int main() {
    int T;
    cin >> T;
    while(T--) {
        cin >> n >> m;
        for(int i=0; i<n; i++) {
            scanf("%d",&a[i]);
        }
        for(int i=0; i<m; i++) {
            scanf("%d",&b[i]);
        }
        get_nextval();
        int temp = kmp_index(0);
        if(temp != -1)
            ++ temp;
        cout << temp << endl;
    }
    return 0;
}

2264: sequence

Time Limit: 1 Sec Memory Limit: 128 MB
Commits: 578 Resolved: 104
[ Commit ][ Status ][ Discussion Board ][Assert By: admin ]

Topic description

Given a sequence A with n numbers and a sequence B with m (m<=n) numbers.

Ask how many consecutive subsequences of length m A _k ,A _k+1 ,...,A _k+m-1 exist in A such that for any 1<=i, j<=m satisfies _Ak+i-1 -B _i =A _k+j-1 -B _j

enter

The first line contains two integers n, m (1 <= m <= n <= 10 ⁶ )

The next line contains n integers describing the sequence A (A _i <= 10 ⁹ )

The next line contains m integers describing the sequence B (B _i <= 10 ⁹ )

output

output a number to indicate the answer

sample input

7 4

6 6 8 5 5 7 4

7 7 9 6

Sample output

2

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int N = 1000100;
int n, m;
int a[N], b[N];
int nextt[N];

void get_nextt()
{
    int i = 0, j;
    j = nextt[0] = -1;
    while(i < m)
    {
        if(-1 == j || b[i] == b[j] )
            nextt[++i] = ++j;
        else
            j = nextt[j];
    }
}

int kmp_count(int pos) {
    int i = pos, j = 0;
    int years = 0;

    while(i < n) {
        if(-1 == j || a[i] == b[j]) {
            ++i, ++j;
            if(j >= m) {
                ++ years;
                j = nextt[j];
            }
        }
        else
            j = nextt[j];
    }
    return ans;
}

int main() {

    cin >> n >> m;
    for(int i=0; i<n; i++)
        scanf("%d",&a[i]);
    for(int i=0; i<m; i++)
        scanf("%d",&b[i]);
    n--, m--;
    for(int i=0; i<n; i++)
        a[i] = a[i+1] - a[i];
    for(int i=0; i<m; i++)
        b[i] = b[i+1] - b[i];
   /* for(int i=0; i<m; i++) {
        cout << nextt[i] << " ";
    }
    cout << endl;*/
    get_nextt();
    int ans = kmp_count(0);
    cout << ans << endl;
    return 0;
}