hdoj 1711 Number Sequence 【KMP】

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 15101    Accepted Submission(s): 6623

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

#include<stdio.h>#include<string.h>#include<stdlib.h>const int N=1000000+10;const int M=10000+10;int a[N],b[M];int next[N];int ls,lm;void getn(){int i = 1,j = 0;next[0] = -1;next[1] = 0;while(i < ls){if(j == -1||b[i] == b[j]){i++,j++;next[i] = j;  //求出 子串 的模式串 }elsej = next[j];}}int kmp(){int i,j;getn();i = 0;j = 0;while(i < lm&&j != ls){if(j == -1||a[i]==b[j]) //逐位比较子串与母串 {i++,j++;}elsej = next[j];}if( j == ls )return i-ls+1; //从0开始所以要再加上 1. elsereturn -1; //不满足的话 输出 -1. }int main(){int n,i,j,k;scanf("%d",&n);while(n--){scanf("%d%d",&lm,&ls);for(i = 0;i < lm; i++)scanf("%d",&a[i]);for(j = 0;j < ls;j++)scanf("%d",&b[j]);k = kmp();  printf("%d\n",k);}return 0;}