HDU1711 Number Sequence(字符串匹配KMP)

题目：

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26616    Accepted Submission(s): 11229

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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Statistic | Submit | Discuss | Note思路：

KMP算法模板题,求子串首次出现的位置

代码：

#include<cstdio>#include<cstring>#include<cctype>#include<string>#include<set>#include<iostream>#include<stack>#include<cmath>#include<queue>#include<vector>#include<algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10000007#define debug() puts("what the fuck!!!")#define N 1000020#define ll longlongusing namespace std;int a[N],b[10020],nex_t[10020];int n,m;void get_next()//构建next数组{    int j=0,k=-1;    nex_t[0]=-1;    while(j<m)    {        if(k==-1||b[j]==b[k])            nex_t[++j]=++k;        else            k=nex_t[k];    }}int KMP_index()//返回首次出现位置{    int i=0,j=0;    get_next();    while(i<n&&j<m)    {        if(j==-1||a[i]==b[j])        {            i++,j++;        }        else            j=nex_t[j];    }    if(j==m)        return i-m+1;    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        for(int i=0;i<m;i++)            scanf("%d",&b[i]);        printf("%d\n",KMP_index());    }    return 0;}