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
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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 ",kmp()); 60 } 61 }