KMP算法
KMP的基处题目，数字数组的KMP算法应用。
主要是模式串的处理，当模式串内有重复时，模式串向左回溯重复的点的位置（next[]）。
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
1 #include<stdio.h>
2 int a[1000001],b[10001],next[10001];
3 void getnext(int m){
4 int i=1,j=0;
5 next[1]=0;
6 while(i<m){
7 if(j==0||b[i]==b[j]){
8 i++; j++; next[i]=j;
9 }
10 else j=next[j];
11 }
12 }
13
14 void getk(int n,int m){
15 int i=1,j=1;
16 while(i<=n&&j<=m){
17 if(j==0||a[i]==b[j]){i++; j++;}
18 else j=next[j];
19 }
20 if(j>m) printf("%d\n",i-m);
21 else printf("-1\n");
22 }
23
24 int main()
25 {
26 int t,n,m,i,j;
27 scanf("%d",&t);
28 while(t--){
29 scanf("%d%d",&n,&m);
30 for(i=1;i<=n;i++) scanf("%d",&a[i]);
31 for(i=1;i<=m;i++) scanf("%d",&b[i]);
32 getnext(m);
33 getk(n,m);
34 }
35 return 0;
36 }