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.
InputThe 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].
OutputFor 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 首先是预处理Next数组，初始化Next[0]=-1,j=-1 然后从i=0开始，如果j!=-1&&不匹配，j=Next[j]。否则判断p[++i]==p[++j],如果相等，直接让Next[i]=Next[j]，不等的话，让Next[i]=j; 然后在主串里匹配。i=0,j=0; j!=-1&&不匹配， j=Next[j]。否则i++，j++
1 #include<stdio.h>
2 #include<string.h>
3 int Next[10010],t[1000010],p[10010];//模式串对应的Next
4 int n,m,_;
5
6 void prekmp() {//预处理Next数组
7 int i,j;
8 j=Next[0]=-1;
9 i=0;
10 while(i<m) {
11 while(j!=-1&&p[i]!=p[j]) j=Next[j];
12 if(p[++i]==p[++j]) Next[i]=Next[j];//会快一点 博客体现了这一点
13 else Next[i]=j;
14 }
15 }
16
17 int kmp() {//查找在主串的位置
18 int i=0,j=0;
19 prekmp();
20 while(i<n&&j<m) {
21 while(j!=-1&&t[i]!=p[j]) j=Next[j];
22 i++;j++;
23 }
24 if(j==m) return i-m+1;
25 else return -1;
26 }
27
28 int main() {
29 for(scanf("%d",&_);_;_--) {
30 scanf("%d%d",&n,&m);
31 for(int i=0;i<n;i++) {
32 scanf("%d",&t[i]);
33 }
34 for(int i=0;i<m;i++) {
35 scanf("%d",&p[i]);
36 }
37 int ans=kmp();
38 printf("%d\n",ans);
39 }
40 }
原文地址：https://www.cnblogs.com/ACMerszl/p/10263394.html