個人心得：樸素代碼絕對超時，所以要用到KMP算法，特意了解了，還是比較抽象，要多體會

Given two sequences of numbers : a11, a22, ...... , aNN, and b11, b22, ...... , bMM?(1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aKK?= b11, aK+1K+1?= b22, ...... , aK+M?1K+M?1?= bMM. If there are more than one K exist, output the smallest one.?

InputThe 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 a11, a22, ...... , aNN. The third line contains M integers which indicate b11, b22, ...... , bMM. All integers are in the range of??1000000,1000000?1000000,1000000.?
OutputFor 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 1 #include<stdio.h> 2 #include<string.h> 3 #include<iostream> 4 #include<algorithm> 5 #include<utility> 6 #include<queue> 7 #include<set> 8 using namespace std; 9 int n,m; 10 int a[1000005],b[10005]; 11 void getnext(int x[],int next[]) 12 { 13 int i=0; 14 int j=-1; 15 next[i]=-1; 16 while(i<m) 17 { 18 if(j==-1||x[i]==x[j]) 19 { 20 i++; 21 j++; 22 next[i]=j; 23 } 24 else 25 j=next[j]; 26 } 27 } 28 int main() 29 { 30 int t; 31 scanf("%d",&t); 32 while(t--) 33 { 34 int next[10005]; 35 int flag=-1; 36 scanf("%d%d",&n,&m); 37 for(int i=1;i<=n;i++) 38 scanf("%d",&a[i]); 39 for(int i=0;i<m;i++) 40 scanf("%d",&b[i]); 41 getnext(b,next); 42 int i=1,j=0; 43 while(i<=n&&j<m) 44 { 45 if(j==-1||a[i]==b[j]) 46 { 47 i++; 48 j++; 49 } 50 else 51 j=next[j]; 52 } 53 if(j==m) flag=i-j; 54 cout<<flag<<endl; 55 56 57 } 58 return 0; 59 60 61 }

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轉載于:https://www.cnblogs.com/blvt/p/7270543.html

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