HDU 1711 Number Sequence 【KMP】

题目链接

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

思路：简单KMP模版

WA： xt数组要初始化xt[1] = 1

AC代码：

#include <iostream>#include <stdio.h>#include <cstring>#define N 1000010#define M 10010using namespace std;int a[N],b[M],n,m,xt[M];void getnext(){    int j = 1;    xt[1]=1;//WA的地方！没有加这个初始化错了三次    for(int i = 2; i <= m; i++)    {        while(b[i] != b[j] && j > 1)            j = xt[j-1];        if(b[i] == b[j])            j++;        xt[i] = j;    }}int KMP(){    if(n < m)        return -1;    int j = 1;    for(int i = 1; i <= n; i++)    {        while(a[i] != b[j] && j > 1)            j = xt[j-1];        if(a[i] == b[j])            j++;        if(j == m+1)            return i-m+1;    }    return -1;}int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d%d",&n, &m);        for(int i = 1; i <= n; i++)            scanf("%d", &a[i]);        for(int i = 1; i <= m; i++)            scanf("%d", &b[i]);        getnext();        printf("%d\n", KMP());    }}