HDU 1711 Number Sequence(KMP)

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27986 Accepted Submission(s): 11777

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

KMP，返回第一次出现的地址。只是不是字符串而是int数组而已。

#include <stdio.h>#include <iostream>#include <string.h>using namespace std;int x[1000000+5];int y[10000+5];int l1,l2;int nxt[10000+5];void kmpnext(){    int i=0,j=-1;    nxt[0]=-1;    while(i<l2)    {        while(j!=-1&&y[i]!=y[j])            j=nxt[j];        nxt[++i]=++j;    }    }int kmp(){    kmpnext();    int i=0,j=0;    while(i<l1)    {        while(j!=-1&&x[i]!=y[j])            j=nxt[j];        i++;j++;        if(j>=l2)        {            return i-l2+1;        }    }    return -1;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(nxt,0,sizeof nxt);        scanf("%d %d",&l1,&l2);        for(int i=0;i<l1;i++)        {            scanf("%d",&x[i]);        }        for(int i=0;i<l2;i++)        {            scanf("%d",&y[i]);        }        printf("%d\n",kmp());            }}

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