HDU 1711 Number Sequence(KMP)
2014-03-17 14:55
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Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9548 Accepted Submission(s): 4370
Problem Description
Given two sequences of numbers : a[1], a[2], ...... , a
, 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
.
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: 求最先匹配的位置。
思路:第一次做KMP,这题裸题,熟悉一下。
代码:
#include <stdio.h> #include <string.h> const int N = 1000005; const int M = 10005; int n, m, seq , seq1[M], next[M], i, t; void get_next(int *seq, int m) { memset(next, 0, sizeof(next)); int j = 0; for (int i = 2; i <= m; i++) { while (j > 0 && seq[i] != seq[j + 1]) j = next[j]; if (seq[i] == seq[j + 1]) j++; next[i] = j; } } int kmp(int *seq, int *seq1, int n, int m) { int j = 0, ans = 0; for (int i = 1; i <= n; i++) { while (j > 0 && seq[i] != seq1[j + 1]) j = next[j]; if (seq[i] == seq1[j + 1]) j++; if (j == m) { return i - m + 1; } } return -1; } int main() { scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); for (i = 1; i <= n; i++) scanf("%d", &seq[i]); for (i = 1; i <= m; i++) scanf("%d", &seq1[i]); get_next(seq1, m); printf("%d\n", kmp(seq, seq1, n, m)); } return 0; }
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