HDU - 1711Number SequenceKMP 模板题

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

题意：给两个数组，要找一个最小的K使得a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]

思路：直接套KMP

#include <iostream> #include<stdio.h> #include<string> #include<string.h> #include<algorithm> using namespace std; #define MAXN 1000010 int p[MAXN], t[10010]; int Next[MAXN]; int m, n; void getNext(int* P, int* f) { f[0] = f[1] = 0; for (int i = 1; i < n; i++) { int j = f[i]; while (j&&P[i] != P[j]) { j = f[j]; } f[i + 1] = P[j] == P[i] ? j + 1 : 0; } } int KMP(int* T, int* P, int* f) { int cnt = MAXN; getNext(P, f); int j = 0; for (int i = 0; i < m; i++) { while (j&&P[j] != T[i])j = f[j]; if (P[j] == T[i])j++; if (j == n) { cnt = min(i - j + 2, cnt); break; } } return cnt; } int main() { int T; while (~scanf("%d", &T)) { while (T--) { scanf("%d %d", &m, &n); for(int i=0;i<m;i++){ scanf("%d", &p[i]); } for (int i = 0; i < n; i++) { scanf("%d", &t[i]); } int minn= KMP(p, t, Next); if (minn == MAXN) { printf("-1\n"); } else { printf("%d\n", minn); } } } }