HDU 1711 Number Sequence【KMP】【模板题】【水题】（返回匹配到的第一个字母的位置）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29634    Accepted Submission(s): 12464

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

返回匹配到的第一个字母的位置

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <stdlib.h>#include <string>#include <cstring>#include <map>#include <set>#include <queue>#include <stack>#define INF 0x3f3f3f3f#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;const double pi = acos(-1.0);const int mod = 1e9 + 7;const int maxn = 1e5 + 10;int nextval[1000010];int s[1000010], p[10010];int slen, plen;//p为模式串void getnext(int p[], int nextval[])      //朴素kmp，nextval[i]即为1~i-1的最长前后缀长度{    int len = plen;    int i = 0, j = -1;    nextval[0] = -1;    while (i < len)    {        if (j == -1 || p[i] == p[j])        {            nextval[++i] = ++j;        }        else            j = nextval[j];    }}//在s中找p出现的位置int KMP(int s[], int p[], int nextval[]){    getnext(p, nextval);    int ans = 0;    int i = 0;  //s下标    int j = 0;  //p下标    int s_len = slen;    int p_len = plen;    while (i < s_len && j < p_len)    {        if (j == -1 || s[i] == p[j])        {            i++;            j++;        }        else            j = nextval[j];        if (j == p_len)        {            return i - j + 1;        }    }    return -1;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d%d", &slen, &plen);        for (int i = 0; i < slen; i++) scanf(" %d", &s[i]);        for (int i = 0; i < plen ; i++) scanf(" %d", &p[i]);        printf("%d\n", KMP(s, p, nextval));    }    return 0;}