HDU 1711 Number Sequence(KMP裸题，板子题，有坑点)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27028    Accepted Submission(s): 11408

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

分析：KMP裸题，自己看吧，不会的看我博客详解！此题有道坑点就是读入不能用cin读入，很容易T！

纯粹要看运气才会过QAQ

优化以后：

速度快了将近3.5s，scanf大法好啊

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 const int N=;
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 int kmpnext[N];
 int s[N],t[N];///s为主串，t为模式串
 int slen,tlen;///slen为主串的长度，tlen为模式串的长度
 inline void getnext()
 {
     int i,j;
     j=kmpnext[]=-;
     i=;
     while(i<tlen)
     {
         if(j==-||t[i]==t[j])
         {
             kmpnext[++i]=++j;
         }
         else
         {
             j=kmpnext[j];
         }
     }
 }
 /*
 返回模式串T在主串S中首次出现的位置
 返回的位置是从0开始的。
 */
 inline int kmp_index()
 {
     int i=,j=;
     getnext();
     while(i<slen&&j<tlen)
     {
         if(j==-||s[i]==t[j])
         {
             i++;
             j++;
         }
         else
             j=kmpnext[j];
     }
     if(j==tlen)
         return i-tlen;
     else
         return -;
 }
 /*
 返回模式串在主串S中出现的次数
 */
 inline int kmp_count()
 {
     int ans=;
     int i,j=;
     if(slen==&&tlen==)
     {
         if(s[]==t[])
             return ;
         else
             return ;
     }
     getnext();
     for(i=;i<slen;i++)
     {
         while(j>&&s[i]!=t[j])
             j=kmpnext[j];
         if(s[i]==t[j])
             j++;
         if(j==tlen)
         {
             ans++;
             j=kmpnext[j];
         }
     }
     return ans;
 }
 int T;
 int main()
 {
     T=read();
     while(T--)
     {
         slen=read();
         tlen=read();
         for(int i=;i<slen;i++)
             s[i]=read();
         for(int i=;i<tlen;i++)
             t[i]=read();
         if(kmp_index()==-)
             cout<<-<<endl;
         else
             cout<<kmp_index()+<<endl;
     }
     return ;
 }