【2018.07.27】（字符串/找相同）学习KMP算法小记

虽然说原理很好理解，但是代码理解了花费我一个下午的时间，脑阔痛

该注释的地方都标记了，希望以后看到这些代码我还能好好理解吧

学习的链接地址：https://www.cnblogs.com/teble/p/7280575.html

/*     Number Sequence     */
/*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*/
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#define MAXN 1000050
using namespace std;
//这道题看得出来需要使用到KMP排序
//虽然说我理解了排序原理，却不太懂得实现它
//对于我这种来说，最担心的的莫过于输入问题（所以最后干脆用int数组解决）
int a[MAXN], b[], _next[];
int repeat, i, n;
__int64 m;

void makeNext()
{
    int q=;//模板b字符串的临时下标，作为比较的后一位
    int k=;//最大前后缀长度
    _next[]=;//模板字符串第一个字符最大前后缀长度为0
    while ( q<n )//遍历数组
    {
        if ( k== || b[k]==b[q] )//符合条件
        {
            ++k,++q;//同时+1，判断下一位
            if ( b[q]!=b[k] ) _next[q]=k;
                else _next[q]=_next[k];//上下两句话改变next数组的值
        } else k=_next[k];//如果不相等，将前标往前移动
    }
    /*数据检查for ( k=1 ; k<=n ; k++ )
    {
        cout<<b[k]<<' ';
    }
    cout<<endl;
    for ( k=1 ; k<=n ; k++ )
    {
        cout<<_next[k]<<' ';
    }
    cout<<endl;*/
}

int kmp()
{
    int i, j;
    for ( i=,j= ; i<=m&&j<=n ; )
    {
        if ( j== || a[i]==b[j] ) ++i,++j;
            else j=_next[j];//如果不相等的话，将此时的j变小，相当于将b数组往右移动
            //next数组提供，当a[]的后m个元素和b[]开头的前m个元素顺序相同时提供下标的服务
    }
    if ( j==n+ ) return i-n;//i是当前点，n是b[]的长度，返回的是初始点
        else return -;//j==n+1的原因是a[]都被遍历完了，而b[]仍未遍历完
        //只有当b[]都被遍历完了才算完整结束
}

int main(void)
{
    scanf("%d", &repeat);
    while ( repeat-- )
    {
        scanf("%I64d%d",&m ,&n);
        for ( i= ; i<=m ; i++ )
        {
            scanf("%d", &a[i]);
        }
        for ( i= ; i<=n ; i++ )
        {
            scanf("%d", &b[i]);
        }
        makeNext();
        cout<<kmp()<<endl;
    }
    return ;
}