HDOJ 1711 Number Sequence ——KMP

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11789    Accepted Submission(s): 5380

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

 

纯KMP模板：

#include <stdio.h>int ms[100010];int next[100010];int zc[1000010];void getnext(int lenms){    int i = 0,j = -1;    next[i] = j;    while(i < lenms)    {        if(j == -1 || ms[i] == ms[j])        {            i++;            j++;            next[i] = j;        }        else        {            j = next[j];        }    }}void KMP(int lenms,int lenzc){    int i = 0,j = 0;    while(i < lenzc && j < lenms)    {        if(j == -1 || zc[i] == ms[j])        {            i++;            j++;        }        else        {            j = next[j];        }    }    if(j >= lenms)        printf("%d\n",i - j + 1);    else        printf("-1\n");}int main(){    int t,lenms,lenzc;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&lenzc,&lenms);        for(int i = 0;i < lenzc;i++)        {            scanf("%d",&zc[i]);        }        for(int i = 0;i < lenms;i++)        {            scanf("%d",&ms[i]);        }        getnext(lenms);        KMP(lenms,lenzc);    }    return 0;}