Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30591 Accepted Submission(s): 12870

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
5
2 1 2 3 1 2 3 1 3 2 1 2
2 3 1 3
5
2 1 2 3 1 2 3 1 3 2 1 2
2 3 2 1

Sample Output

6
-1

Source

HDU 2007-Spring Programming Contest

把数字离散化后当作字符来处理就是kmp了，注意模板是从下标0开始的我一直输入时从1开始导至答案不对。

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<map>
 4 #include<iostream>
 5 using namespace std;
 6 map<int,int>M;
 7 int nex[10005],l1,l2,sl;
 8 int S[1000005],T[10005];
 9 int solve()
10 {
11     int i,j;
12     nex[0]=nex[1]=0;
13     for(i=1;i<l2;++i)
14     {
15         j=nex[i];
16         while(j&&T[i]!=T[j])j=nex[j];
17         nex[i+1]=T[i]==T[j]?j+1:0;
18     }
19     j=0;
20     for(i=0;i<l1;++i)
21     {
22         while(j&&S[i]!=T[j])j=nex[j];
23         if(S[i]==T[j]){
24             j++;
25             if(j==l2)return i-l2+2;
26         }
27     }
28     return -1;
29 }
30 int main()
31 {
32     int i,j,t,p,x,tmp;
33     cin>>t;
34     while(t--){p=0;M.clear();
35         scanf("%d%d",&l1,&l2);
36         for(i=0;i<l1;++i)
37         {
38             scanf("%d",&x);
39             tmp=M[x];
40             if(!tmp){M[x]=S[i]=++p;}
41             else S[i]=tmp;
42         }
43         for(i=0;i<l2;++i)
44         {
45             scanf("%d",&x);
46             tmp=M[x];
47             if(!tmp){M[x]=T[i]=++p;}
48             else T[i]=tmp;
49         }
50         cout<<solve()<<endl;;
51     }
52     return 0;
53 }