HDU 1711 Number Sequence

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16243 Accepted Submission(s): 7165

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

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
KMP算法入门题目，给出主串和模式串以及各自的长度，问模式串在主串中第一次出现的位置，找到输出下标，找不到输入-1
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1000000+5;int S[maxn];int T[10005];int Nextval[maxn];int n,m;void getNextval(){int i=1;Nextval[1]=0;int j=0;while(i<m){if(j==0 || T[i]==T[j]){i++,j++;if(T[i]!=T[j]) Nextval[i] = j;else Nextval[i] = Nextval[j];}else j = Nextval[j];}}int merge(){int i,j,k,t;i = 1,j = 1;while(i<=n && j<=m){if(j==0 || S[i]==T[j]) { i++, j++; }else            j = Nextval [j];}if(j>m) return i-m;else return -1;}int main(){    int i,j,t;cin>>t;while(t--){cin>>n>>m;for(i=1;i<=n;i++)            scanf("%d",&S[i]);for(i=1;i<=m;i++)scanf("%d",&T[i]);        getNextval();cout<<merge()<<endl;}return 0;}

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