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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
Source
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题解:kmp模板 编译错误了多次,无爱啦...
代码:
1 #include <iostream>
2 #include <cstdio>
3 using namespace std;
4 int p[10010],s[1000010];
5 int n,m;
6 int nex[10010];
7
8 void get()
9 {
10 int plen=m;
11 nex[0]=-1;
12 int k=-1,j=0;
13 while(j < plen){
14 if(k==-1 || p[j] == p[k]){
15 ++j;
16 ++k;
17 if(p[j] != p[k])
18 nex[j]=k;
19 else
20 nex[j]=nex[k];
21 }
22 else{
23 k=nex[k];
24 }
25 }
26 }
27
28 int kmp()
29 {
30 int i=0,j=0;
31 int slen=n;
32 int plen=m;
33 while(i < slen && j< plen){
34 if(j==-1 || s[i]==p[j]){
35 ++i;
36 ++j;
37 }
38 else{
39 j=nex[j];
40 }
41 }
42 if(j == plen)
43 return i-j+1;
44 else
45 return -1;
46 }
47
48 int main()
49 {
50 int t;
51 scanf("%d",&t);
52 while(t--){
53 scanf("%d%d",&n,&m);
54 for(int i=0; i<n; i++)
55 scanf("%d",&s[i]);
56 for(int i=0; i<m; i++)
57 scanf("%d",&p[i]);
58 get();
59 printf("%d\n",kmp());
60 }
61 }
1711 Number Sequence(kmp)
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原文地址：http://www.cnblogs.com/wangmengmeng/p/5487174.html
