Number Sequence（KMP）

Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29708 Accepted Submission(s): 12504

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

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

本题的题意就是查看b串是不是a串的子串 如果是的话返回最小的下标 如果不是的话返回-1 值得注意的是本题最好用scanf和printf进行输入输出 如果用cin和cout时间会超时

#include<iostream>#include<string>#include<string.h>using namespace std;int target[1000007], pattern[10007];int nxt[10007];int lens,lenp;void Make_Next(){    int i = 0,j=-1;    nxt[0] = -1;    while(i!=lenp)    {        if(j==-1||pattern[i]==pattern[j])            nxt[++i] = ++j;        else            j = nxt[j];    }}int kmp(){    int i =0,j = 0;    while(i!=lens&&j!=lenp)    {        if(j==-1||target[i]==pattern[j])            i++,j++;        else            j = nxt[j];         if(j>=lenp)        {            printf("%d\n",i-j+1);            return 0;        }    }   printf("-1\n");    return 0;}int main(){    int ncase;    cin>>ncase;    while(ncase--)    {        scanf("%d%d",&lens,&lenp);        for(int i = 0;i<lens;i++)            scanf("%d",&target[i]);        for(int j = 0;j<lenp;j++)          scanf("%d",&pattern[j]);        Make_Next();        kmp();    }    return 0;}