杭电 1711 KMP算法

         是一道KMP的水题，，，裸题，，，就是一个简单的模板应用。。。。。。题目：

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3689    Accepted Submission(s): 1688

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
   
   
   
   

 

ac代码：用的G++，跑了1000多ms，，，求优化。

#include <iostream>
#include <string>
#include <cstdio>
#include <string.h>
using namespace std;
const int N=1000005,M=100005;
int a[N],b[M],nextt[M];
int n,m;
void getnext(){
  int i=1,j=0;
  nextt[1]=0;
  while(i<m){
	  if(j==0||b[i]==b[j]){
	    ++i;++j;
		if(b[i]!=b[j])
			nextt[i]=j;
		else
			nextt[i]=nextt[j];
	  }
	  else
		  j=nextt[j];
  }
}
int kmp(){
  int i=1,j=1;
  while(i<=n&&j<=m){
	  if(j==0||a[i]==b[j]){
	    ++i;++j;
	  }
	  else
		  j=nextt[j];
  }
  if(j>m)
	  return i-m;
  return 0;
}
int main(){
	int kk;
	scanf("%d",&kk);
	while(kk--){
	  memset(a,0,sizeof(a));
	  memset(b,0,sizeof(b));
	  memset(nextt,0,sizeof(nextt));
	  //int n,m;
	  scanf("%d%d",&n,&m);
	  for(int i=1;i<=n;++i)
		  scanf("%d",&a[i]);
	  for(int i=1;i<=m;++i)
		  scanf("%d",&b[i]);
	  getnext();
	  int flag=kmp();
	  if(flag==0)
		  printf("-1\n");
	  else
		  printf("%d\n",flag);
	}
  return 0;
}