KMP ( Number Sequence )

最新推荐文章于 2021-03-01 20:47:37 发布

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本文介绍了一种使用KMP算法解决两个数列匹配问题的方法。具体问题为：给定两个整数序列，找到第一个序列中与第二个序列完全匹配的起始位置，若有多个匹配则输出最小的起始位置。通过KMP算法实现高效查找。

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题目链接：https://cn.vjudge.net/contest/248066#problem/A

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

KMP 算法

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 1e6+100;
int nxt[MAXN];

int  s1[MAXN];
int  s2[MAXN];

int n,m;
void getnxt(){
    int i=1;
    int j=0;
    nxt[0]=0;
    while(i<n)
    {
        if(s1[i]==s1[j]){
            nxt[i]=j+1;
            i++;
            j++;
        }else if(!j){
            i++;
        }else{
            j=nxt[j-1];
        }
    }
}
int kmp(){

    int i=0;
    int j=0;
    while(j<n&& i<m){

        if(s1[j]==s2[i]){
            j++;
            i++;
        }else if(!j){
            i++;
        }else{
            j=nxt[j-1];
        }
        if(j==n){
            return i-n+1;
        }
    }
    return -1;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&m,&n);

        for(int i=0;i<m;i++){
            scanf("%d",&s2[i]);
        }

        for(int i=0;i<n;i++){
            scanf("%d",&s1[i]);
        }

        getnxt();

        cout<<kmp()<<endl;
    }


    return 0;
}