【HDU 1711】Number Sequence 【KMP 模板】

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

代码

#include<cstring>#include<cstdio>using namespace std ;typedef long long LL ;const int MAXN = 1000000+10 ;const int MAXM = 1e5 ;const int mod  = 1e9+7;int s[1000000+10],t[10000+10];int n,m;int nexts[10000+10];void get(){    int i,j;i=j=0;    nexts[0]=-1;j=-1;    while(i<m){        if(j==-1||t[i]==t[j])  nexts[++i]=++j;        else j=nexts[j];    }}int Kmp(){    int i,j;i=j=0;    while(i<n){        if(j==-1||s[i]==t[j]) {            ++i;++j;            if(j==m) return i-m+1;        }else j=nexts[j];    }    return -1;}int main(){     int z; scanf("%d",&z);     while(z--){        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++) scanf("%d",&s[i]);        for(int i=0;i<m;i++) scanf("%d",&t[i]);        get();        printf("%d\n",Kmp());    }    return  0;}