hdu 1711 Number Sequence (KMP)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28879    Accepted Submission(s): 12146

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

 

Source

HDU 2007-Spring Programming Contest

 

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lcy

题意：找到第一个匹配的下标。

裸的KMP，直接代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t[1000005],p[10005],ne[10005];void makeNext(const int p[],int ne[],int len){    ne[0]=0;    for(int i=1,k=0;i<len;i++)    {        while(k>0&&p[i]!=p[k]) k=ne[k-1];        if(p[i]==p[k]) k++;        ne[i]=k;    }}int kmp(const int t[],const int p[],int ne[],int n,int m){    int ans=0;    makeNext(p,ne,m);    for(int i=0,k=0;i<n;i++)    {        while(k>0&&t[i]!=p[k]) k=ne[k-1];        if(t[i]==p[k]) k++;        if(k==m)        {            ans=i-m+2;            return ans;        }    }    return -1;}int main(){    int c;    scanf("%d",&c);    for(int i=1;i<=c;i++)    {        int n,m;        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)            scanf("%d",&t[i]);        for(int i=0;i<m;i++)            scanf("%d",&p[i]);        printf("%d\n",kmp(t,p,ne,n,m));    }    return 0;}