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
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：
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int N=1000050; 4 inline int read() 5 { 6 int x=0,f=1; 7 char ch=getchar(); 8 while(ch<'0'||ch>'9') 9 { 10 if(ch=='-') 11 f=-1; 12 ch=getchar(); 13 } 14 while(ch>='0'&&ch<='9') 15 { 16 x=x*10+ch-'0'; 17 ch=getchar(); 18 } 19 return x*f; 20 } 21 int kmpnext[N]; 22 int s[N],t[N];///s为主串，t为模式串 23 int slen,tlen;///slen为主串的长度，tlen为模式串的长度 24 inline void getnext() 25 { 26 int i,j; 27 j=kmpnext[0]=-1; 28 i=0; 29 while(i<tlen) 30 { 31 if(j==-1||t[i]==t[j]) 32 { 33 kmpnext[++i]=++j; 34 } 35 else 36 { 37 j=kmpnext[j]; 38 } 39 } 40 } 41 /* 42 返回模式串T在主串S中首次出现的位置 43 返回的位置是从0开始的。 44 */ 45 inline int kmp_index() 46 { 47 int i=0,j=0; 48 getnext(); 49 while(i<slen&&j<tlen) 50 { 51 if(j==-1||s[i]==t[j]) 52 { 53 i++; 54 j++; 55 } 56 else 57 j=kmpnext[j]; 58 } 59 if(j==tlen) 60 return i-tlen; 61 else 62 return -1; 63 } 64 /* 65 返回模式串在主串S中出现的次数 66 */ 67 inline int kmp_count() 68 { 69 int ans=0; 70 int i,j=0; 71 if(slen==1&&tlen==1) 72 { 73 if(s[0]==t[0]) 74 return 1; 75 else 76 return 0; 77 } 78 getnext(); 79 for(i=0;i<slen;i++) 80 { 81 while(j>0&&s[i]!=t[j]) 82 j=kmpnext[j]; 83 if(s[i]==t[j]) 84 j++; 85 if(j==tlen) 86 { 87 ans++; 88 j=kmpnext[j]; 89 } 90 } 91 return ans; 92 } 93 int T; 94 int main() 95 { 96 T=read(); 97 while(T--) 98 { 99 slen=read(); 100 tlen=read(); 101 for(int i=0;i<slen;i++) 102 s[i]=read(); 103 for(int i=0;i<tlen;i++) 104 t[i]=read(); 105 if(kmp_index()==-1) 106 cout<<-1<<endl; 107 else 108 cout<<kmp_index()+1<<endl; 109 } 110 return 0; 111 }