A - Number Sequence
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 1711
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
题目大意：
给你一个长度为n和长度为m的串，然后让你求出这个长度为m的串第一次在长度为n的串中出现的下标是什么（当然是完全匹配后的），如果有
多个地方出现了这个长度为m的串，那我们就输出最小的下标编号。
解题思路：
直接上kmp，先对模式串通过O(m)的时间复杂度，计算出next[]，然后通过kmp找到找到第一次匹配成功的模式串在文本串中的位置，然后输出
t1-len2+1就OK了。
代码：
1 # include<cstdio>
2 # include<iostream>
3 # include<cstring>
4
5 using namespace std;
6
7
8 int a[1000004];
9 int b[10004];
10 int nxt[10004];
11 int n,m;
12
13 void get_next()
14 {
15 int len = m;
16 int t1 = 0, t2;
17 t2 = nxt[0] = -1;
18 while ( t1<len )
19 {
20 if ( t2==-1||b[t1]==b[t2] )
21 {
22 t1++;t2++;
23 nxt[t1] = t2;
24 }
25 else
26 t2 = nxt[t2];
27 }
28 }
29
30
31 int kmp ( int *a,int *b )
32 {
33 int len1 = n, len2 = m;
34 int t1 = 0,t2 = 0;
35 while ( t1<len1&&t2<len2 )
36 {
37 if ( t2==-1||a[t1]==b[t2] )
38 {
39 t1++;t2++;
40 }
41 else
42 t2 = nxt[t2];
43 }
44 if ( t2==len2 )
45 return t1-t2+1;
46 else
47 return -1;
48 }
49
50
51 int main(void)
52 {
53 int t;scanf("%d",&t);
54 while ( t-- )
55 {
56 scanf("%d%d",&n,&m);
57 for ( int i = 0;i < n;i++ )
58 scanf("%d",&a[i]);
59 for ( int j = 0;j < m;j++ )
60 scanf("%d",&b[j]);
61 get_next();
62 int ans = kmp(a,b);
63 printf("%d\n",ans);
64 }
65
66
67
68 return 0;
69 }