Time Limit: 5000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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
Source
Hmm………..KMP模板题,char改成int就好。
AC代码:
#include <cstdio> using namespace std; int a[1000010], b[10010]; int n, m; void input() { scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) scanf("%d", &a[i]); for (int i = 0; i < m; i++) scanf("%d", &b[i]); } void kmp_pre(int x[], int m, int next[]) { int i, j; j = next[0] = -1; i = 0; while (i < m) { while (-1 != j && x[i] != x[j]) j = next[j]; next[++i] = ++j; } } int next[10010]; int kmp(int x[], int m, int y[], int n) { int i, j; kmp_pre(x, m, next); i = j = 0; while (i < n) { while (-1 != j && y[i] != x[j]) j = next[j]; i++, j++; if (j == m) return i - j + 1; } return -1; } void solve() { printf("%d\n", kmp(b, m, a, n)); } int main() { int T; scanf("%d", &T); while (T--) { input(); solve(); } return 0; }
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