HDU Problem 1711 Number Sequence 【KMP】

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22458 Accepted Submission(s): 9603

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

Sample Output

6
-1

Source

HDU 2007-Spring Programming Contest

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#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long Long;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1.0);
const int MOD = 1e9 + 5;
const int MAXN = 10000 + 5;
int str1[MAXN*100], str2[MAXN];
int Next[MAXN];
void make_next(int P[], int m) {
    int q, k = 0; Next[0] = 0;
    for (q = 1; q < m; q++) {
        while(k > 0 && P[q] != P[k]) k = Next[k-1];
        if (P[q] == P[k])  k++;
        Next[q] = k;
    }
}
int main() {
    int t, m, n;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++) {
            scanf("%d", &str1[i]);
        }
        for (int i = 0; i < m; i++) {
            scanf("%d", &str2[i]);
        }
        int ans = -1, j = 0;
        make_next(str2, m);
        for (int i = 0; i < n; i++) {
            while (j && str1[i] != str2[j]) j = Next[j - 1];
            if (str1[i] == str2[j]) j++;
            if (j == m) {
                ans = i + 2 - j;
                break;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}