【kmp】hdu1171 Number Sequence-白红宇

【kmp】hdu1171 Number Sequence

阅读量：6264 次

发布时间：2019-06-22

本文共 2609 字，大约阅读时间需要 8 分钟。

Time Limit: 5000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

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
      
        2 #include
       
         3 #include
        
          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

转载于:https://www.cnblogs.com/kiritoghy/p/4689079.html

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