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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9608 Accepted Submission(s): 4392


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

Source HDU 2007-Spring Programming Contest
找模式串在文本串中第一次出現都位置。

//109MS	1252K
#include
#include
int text[1000007],pattern[10007];
int next[10007],n,m;
void pre()
{
    next[0]=-1;
    int j=-1;
    for(int i=1;i=0&&pattern[j+1]!=pattern[i])j=next[j];
        if(pattern[j+1]==pattern[i])j++;
        next[i]=j;
    }
}
int kmp()
{
    int ans=0,j=-1;
    for(int i=0;i=0&&pattern[j+1]!=text[i])j=next[j];
        if(pattern[j+1]==text[i])j++;
        if(j==m-1)return i-m+2;//返回第一次找到的位置
    }
    return -1;//找不到
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i