传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1711
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42917 Accepted Submission(s): 17715
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
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
HDU 2007-Spring Programming Contest
题意概括：
给出一串主串，一串子串；
求成功匹配到子串的最小下标；
解题思路：
KMP的应用，稍微变形，匹配到子串就跳出来输出。
AC code：
1 #include <cstdio>
2 #include <iostream>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #define INF 0x3f3f3f3f
7 using namespace std;
8 const int MAXN = 1e6+10;
9 const int MAXM = 1e4+10;
10 int W[MAXM], T[MAXN];
11 int wlen, tlen;
12 int nxt[MAXM];
13
14 void get_nxt()
15 {
16 int j, k;
17 j = 0;
18 k = -1;
19 nxt[0] = -1;
20 while(j < wlen){
21 if(k == -1 || W[j] == W[k]){
22 nxt[++j] = ++k;
23 }
24 else k = nxt[k];
25 }
26 }
27
28 int KMP_index()
29 {
30 int i = 0, j = 0;
31 get_nxt();
32
33 while( i < tlen && j < wlen){
34 if(j == -1 || T[i] == W[j]){
35 i++;
36 j++;
37 }
38 else j = nxt[j];
39 }
40 if(j == wlen) return i-wlen+1;
41 else return -1;
42 }
43
44 int main()
45 {
46 int T_case, N, M;
47 scanf("%d", &T_case);
48 while(T_case--)
49 {
50 scanf("%d%d", &N, &M);
51 wlen = M, tlen = N;
52 for(int i = 0; i < N; i++)
53 scanf("%d", &T[i]);
54 for(int j = 0; j < M; j++)
55 scanf("%d", &W[j]);
56
57 printf("%d\n", KMP_index());
58 }
59 return 0;
60 }
原文地址：https://www.cnblogs.com/ymzjj/p/10020540.html