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:
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 }