hdu 1711 Number Sequence KMP 基础题

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11691    Accepted Submission(s): 5336

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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题意：给两个数组a和b，求b在a中出现的的第一个位置，若没有则输出-1，只是数组变成了整型数组，方法与字符数组完全一样。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1000005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int s[maxn],p[maxn],nextval[maxn];int N,M;void get_nextval(){    int i=0,j=-1;    nextval[0]=-1;    while (i<M)    {        if (j==-1||p[i]==p[j])        {            i++;            j++;            if (p[i]!=p[j])                nextval[i]=j;            else                nextval[i]=nextval[j];        }        else            j=nextval[j];    }}int KMP(){    int i=0,j=0;    while (i<N&&j<M)    {        if (j==-1||s[i]==p[j])        {            i++;            j++;        }        else            j=nextval[j];    }    if (j==M)        return i-M+1;    else        return -1;}int main(){    int cas;    scanf("%d",&cas);    while (cas--)    {        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",&p[i]);        get_nextval();        int ans=KMP();        printf("%d\n",ans);    }    return 0;}/*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*/