1711 Number Sequence(kmp)

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

 

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题解:kmp模板 编译错误了多次,无爱啦...

代码:

 #include <iostream>
 #include <cstdio>
 using namespace std;
 int p[],s[];
 int n,m;
 int nex[];

 void get()
 {
     int plen=m;
     nex[]=-;
     int k=-,j=;
     while(j < plen){
         if(k==- || p[j] == p[k]){
             ++j;
             ++k;
             if(p[j] != p[k])
                 nex[j]=k;
             else
                 nex[j]=nex[k];
         }
         else{
             k=nex[k];
         }
     }
 }

 int kmp()
 {
     int i=,j=;
     int slen=n;
     int plen=m;
     while(i < slen && j< plen){
         if(j==- || s[i]==p[j]){
             ++i;
             ++j;
         }
         else{
             j=nex[j];
         }
     }
     if(j == plen)
         return i-j+;
     else
         return -;
 }

 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&m);
         for(int i=; i<n; i++)
             scanf("%d",&s[i]);
         for(int i=; i<m; i++)
             scanf("%d",&p[i]);
         get();
         printf("%d\n",kmp());
     }
 }