HDU 1711 Number Sequence（KMP）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25840    Accepted Submission(s): 10919

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

Source

HDU 2007-Spring Programming Contest

 

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KMP裸题，套板子即可

#include<bits/stdc++.h>using namespace std;const int N = 1000000 + 10;int a[N];int b[N];int nex[N];int la,lb;void get_next(){    int i=0,j=-1;    nex[0]=-1;    while(i<lb)    {        if(j==-1||b[i]==b[j])        {            i++;            j++;            nex[i]=j;        }        else            j=nex[j];    }}int kmp(){    get_next();    int i=0,j=0;    while(i<la)    {        if(j==-1||a[i]==b[j])        {            i++;            j++;        }        else            j=nex[j];        if(j==lb)            return i-lb+1;    }    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&la,&lb);        for(int i=0;i<la;i++)            scanf("%d",&a[i]);        for(int i=0;i<lb;i++)            scanf("%d",&b[i]);        printf("%d\n",kmp());    }}