HDU 1711 Number Sequence

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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

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分析：

KMP模板题，学KMP算法的练习题。（其实现在对KMP中的next数组理解的还不是很深，我都是一直理解为next[i]是0—i中的前缀与后缀相同的字符数）

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int A[10005];int B[1000005];int nxt[10005];int n,m,t;int makenext(){    int l,r;    l=-1;r=0;nxt[0]=-1;    while(r<m){        if(l==-1||A[l]==A[r]){            nxt[++r]=++l;        }else{            l=nxt[l];        }    }}int kmp(){   makenext();   int ln=0,lm=0;    while(ln<n&&lm<m){        if(lm==-1||B[ln]==A[lm]){            ++ln;++lm;        }else{            lm=nxt[lm];        }        if(lm>=m)return ln-lm+1;    }    return -1;}int main(){  //  freopen("in.txt","r",stdin);    scanf("%d",&t);    bool dec;    while(t--){        dec=true;        memset(nxt,0,sizeof(nxt));        scanf("%d%d",&n,&m);        for(int i=0;i<n;++i)scanf("%d",&B[i]);        for(int i=0;i<m;++i)scanf("%d",&A[i]);        printf("%d\n",kmp());    }    return 0;}

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