标签：mil position ott space from msu fine return mis
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
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 }
标签：mil position ott space from msu fine return mis
原文地址：http://www.cnblogs.com/haoabcd2010/p/6842448.html
