Number Sequence(C++版）

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

 

不需要什么思路，使用KMP算法找到，出现的位置就行

#include <iostream>
#include <vector>

using namespace std;

const vector<int> getNext(const vector<int>& temp) {
    int i = 1, k = 0;
    vector<int> next(temp.size() + 1);
    next[0] = 0;
    for (; i < temp.size(); ++i) {
        while (k > 0 && temp[k] != temp[i])
            k = next[k - 1];
        if (temp[k] == temp[i])
            ++k;
        next[i] = k;
    }

    return next;
}

int KMP(vector<int> & val1, vector<int> & val2) {
    int k = 0;
    vector<int> next(val2.size() + 1);
    next = getNext(val2);
    for (int i = 0; i < val1.size(); ++i) {
        while (k > 0 && val1[i] != val2[k])
            k = next[k - 1];
        if (val1[i] == val2[k])
            ++k;
        if (k == val2.size())
            return i - k + 2;
    }

    return -1;
}

int main()
{
    int n, i, j;
    cin >> n;
    for (int t = 0; t < n; ++t) {
        cin >> i >> j;
        vector<int> vec1(i), vec2(j);
        for (int m = 0; m < i; ++m) {
            cin >> vec1[m];
        }
        for (int r = 0; r < j; ++r) {
            cin >> vec2[r];
        }

        cout << KMP(vec1, vec2) << endl;
    }

    system("PAUSE");
    return 0;
}