Sequence I (hdu 5918)
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1938 Accepted Submission(s): 730
Problem Description
Mr. Frog has two sequences a1,a2,?,an and b1,b2,?,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,?,bmis exactly the sequence aq,aq+p,aq+2p,?,aq+(m−1)p where q+(m−1)p≤n and q≥1.
Input
The first line contains only one integer T≤100, which indicates the number of test cases.
Each test case contains three lines.
The first line contains three space-separated integers 1≤n≤106,1≤m≤106 and 1≤p≤106.
The second line contains n integers a1,a2,?,an(1≤ai≤109).
the third line contains m integers b1,b2,?,bm(1≤bi≤109).
Output
For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.
Sample Input
2
6 3 1
1 2 3 1 2 3
1 2 3
6
3 2
1
3 2 2 3 1
1
2 3
Sample Output
Case #1: 2
Case #2: 1
//题意: 字符串匹配，就是，n 长主串，m 长匹配串，k 长间隔，问有多少种匹配?
分成 k 组就好，这题可以用来测测你的 KMP 模板哦，数据还可以，就是，就算是朴素匹配也能过。。。
做完我算是真的理解KMP了，对于字符串，有个 \0 结尾的特性，所以优化的 next 是可行的，但是这种却不行，而且，优化的并不好求匹配数。
KMP 模板 : http://www.cnblogs.com/haoabcd2010/p/6722073.html
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <vector>
5 using namespace std;
6 #define MX 1000005
7
8 int n,m,p;
9 int ans;
10 int t[MX];
11 vector<int> zu[MX];
12 int net[MX];
13
14 void Init()
15 {
16 for (int i=0;i<=n;i++)
17 zu[i].clear();
18 }
19
20 void get_next()
21 {
22 int i=0,j=-1;
23 net[0]=-1;
24 while (i<m)
25 {
26 if (j==-1||t[i]==t[j]) net[++i]=++j;
27 else j = net[j];
28 }
29
30 for (int i=0;i<=m;i++)
31 printf("%d ",net[i]);
32 printf("\n");
33 }
34
35 void KMP(int x)
36 {
37 int i=0,j=0;
38 int len = zu[x].size();
39 while(i<len&&j<m)
40 {
41 if (j==-1||zu[x][i]==t[j])
42 {
43 i++,j++;
44 }
45 else j=net[j];
46 if (j==m)
47 {
48 ans++;
49 j = net[j];
50 }
51 }
52 }
53
54 int main()
55 {
56 int T;
57 scanf("%d",&T);
58 for(int cas=1;cas<=T;cas++)
59 {
60 scanf("%d%d%d",&n,&m,&p);
61 Init();
62 for (int i=0;i<n;i++)
63 {
64 int x;
65 scanf("%d",&x);
66 zu[i%p].push_back(x);
67 }
68 for (int i=0;i<m;i++)
69 scanf("%d",&t[i]);
70 get_next();
71
72 ans = 0;
73 for (int i=0;i<p;i++) KMP(i);
74 printf("Case #%d: %d\n",cas,ans);
75 }
76 return 0;
77 }