• kmp


    Number Sequence
    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 42846    Accepted Submission(s): 17687

    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

    题目大意:给出一个母串和一个子串,然后输出子串第一次出现在母串中的位置。

    #include <stdio.h>
    #include <string.h>
    #include<iostream>
    using namespace std;
    const int MAXN = (int) 1e6 + 7;
    int n, m;       // 母串和子串的长度
    int t[MAXN];   // 子串
    int s[MAXN];   // 母串
    int next[MAXN];
    void getNext(int length)
    {
        memset(next, 0, sizeof(next));
        next[0] = -1;
        int j = 0;
        int k = -1;
     
        while(j < length)
        {
            if(k == -1 || t[j] == t[k])
            {
                next[++j] = ++k;
            }
            else
                k = next[k];
        }
    }
    int kmp()
    {
        int i = 0;
        int j = 0;
        while(i < n && j < m)
        {
            if(s[i] == t[j])
                i++, j++;
            else if(j == -1)
                i++;
            else
                j = next[j];
        }
        if(j == m)
            return i - m + 1;
        else
            return -1;
    }
    int main()
    {
        int num;
        scanf("%d", &num);
        while( num-- )
        {
            scanf("%d%d", &n, &m);
            for(int i=0; i<n; i++)
                scanf("%d", &s[i]);
            for(int i=0; i<m; i++)
                scanf("%d", &t[i]);
            getNext(m);
            for(int i=0;i<=m;i++)
                cout<<next[i]<<" ";
            cout<<endl;
            printf("%d
    ", kmp());
     
        }
        return 0;
    }
  • 相关阅读:
    (转)Mat, vector<point2f>,Iplimage等等常见类型转换
    人脸检测与识别的趋势和分析
    经典网络LeNet5看卷积神经网络各层的维度变化
    C++: int int& int * int**的区别、联系和用途
    c++从txt中读取数据,数据并不是一行路径(实用)
    Spyder常用快捷键
    批量读写变换图片(转)
    OpenCV属性页配置问题~
    视觉目标检测和识别之过去,现在及可能(2017.06.28)
    数据库---SQL Server
  • 原文地址:https://www.cnblogs.com/lusiqi/p/12633168.html
Copyright © 2020-2023  润新知