HDU1711-Number Sequence

Number Sequence
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11038    Accepted Submission(s): 5038

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
 

字符串匹配题
KMP算法
AC代码：

#include<stdio.h>#include<string.h>int a[1000010],b[10010],next[10010];int n,m;void getnext(int p[]){    int i = 0;    int j = -1;    next[0] = -1;    while(i < m)    {        if(j == -1 || p[i] == p[j])        {            i++;            j++;            if(p[i]!=p[j])                next[i] = j;            else                next[i] = next[j];        }        else            j = next[j];    }}int KMP(int a[],int b[]){    int j = 0;    int i = 0;    getnext(b);    while(i < n && j < m)    {        if(j == -1 || a[i] == b[j])        {            i++;            j++;        }        else            j = next[j];    }    if(j >= m)        return i - m+1;    else        return -1;}int main(){    int i,j,T;    scanf("%d",&T);    while(T--)    {       scanf("%d %d",&n,&m);       memset(a,0,sizeof(a));       memset(b,0,sizeof(b));       memset(next,0,sizeof(next));       for(i=0;i<n;i++)       scanf("%d",&a[i]);       for(i=0;i<m;i++)       scanf("%d",&b[i]);           printf("%d\n",KMP(a,b));    }    return 0;}