【kmp】hdu1171 Number Sequence

             　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Number Sequence
Time Limit: 5000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u

 

 Status

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

 

 

话说今天才学kmp，于是找了一道裸的kmp的题，写出来发现关于next数组的求解过程还是不太懂。。。居然是靠背的，看来需要多花时间去理解理解，不过好歹还是自己写出来的，放到这来存着吧。。

附一个大神关于kmp的一个超详细讲解，免得以后忘了

【代码】：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 using namespace std;
 5 int T;
 6 int a[1000000 + 5],b[10000 + 5];
 7 int n,m;
 8 
 9 void getnext(int *next)
10 {
11     int k = -1;
12     int j = 0;
13     next[0] = -1;
14     while(j < m)
15     {
16         if(k == -1 || b[j] == b[k])
17         {
18             j++;
19             k++;
20             if(b[j] != b[k]) next[j] = k;
21             else next[j] = next[k];
22         }
23         else k = next[k];
24     }
25 }
26 int kmp(int *next)
27 {
28     int i = 0,j = 0;
29     while(i < n && j < m)
30     {
31         if(j == -1 || a[i] == b[j])
32         {
33             i++;j++;
34         }
35         else j = next[j];
36         if(j == m)return i - j + 1;
37     }
38     return -1;
39 }
40 int main()
41 {
42     int next[10000 + 5];
43     scanf("%d", &T);
44     while(T--)
45     {
46         memset(next,0,sizeof(next));
47         memset(a,0,sizeof(a));
48         memset(b,0,sizeof(b));
49         scanf("%d%d", &n, &m);
50         for(int i = 0; i < n; i++)
51         scanf("%d", &a[i]);
52         for(int i = 0; i < m; i++)
53         scanf("%d", &b[i]);
54         getnext(next);
55         int ans = kmp(next);
56         printf("%d\n",ans);
57     }
58     return 0;
59 }
View Code

想吐槽一下，在oj上提交时，为毛找我的next数组的茬，我开的全局变量好么，一定要我开在main里面么，还非要我传参数

 

以下大神链接

点击打开链接

posted @ 2015-07-30 13:37  kiritoghy  阅读(129)  评论(0)    收藏  举报