HDU-1711

链接:

  http://acm.hdu.edu.cn/showproblem.php?pid=1711

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题目：

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

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

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题意：

  给你a和b两个数组的长度，然后给你a和b数组，让你求b数组在a数组里第一次出现的位置，下标从1开始。

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思路：

  KMP水题。

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实现：

#include <bits/stdc++.h>using namespace std;const int maxn = int(2e7)+7;int str[maxn], p[maxn], Next[maxn], n, m, t;void GetNext() {    for(int q = 1, k = 0 ; q < m ; q++) {        while(k > 0 && p[q] != p[k]) k = Next[k-1];        if(p[q] == p[k]) k++;        Next[q] = k;    }}int KMP() {    GetNext();    for(int i = 0, q = 0 ; i < n ; i++) {        while(q > 0 && p[q] != str[i]) q = Next[q-1];        if(p[q] == str[i]) q++;        if(q == m) return i-m+2;    }    return -1;}int main() {    scanf("%d", &t);    while(Next[0] = 0, t--) {        scanf("%d%d", &n, &m);        for(int i=0 ; i<n ; i++) scanf("%d", str+i);        for(int i=0 ; i<m ; i++) scanf("%d", p+i);        printf("%d\n", KMP());    }    return 0;}