hdu1711 Number Sequence

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10731 Accepted Submission(s): 4879

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

Sample Output

6-1

这是我KMP代码的第一篇，心情好激动。。。

直接贴代码吧：

#include<stdio.h>#include<string.h>int a[1000005];int b[10005];int next[10005];int m,n;void get_next(){    int i,j;    next[1]=0;    j=0;i=1;    while(i<=m)    {        if(j==0||b[i]==b[j]) {i++;j++;next[i]=j;}        else            j=next[j];    }     }int KMP() //注意此函数和get_next可不一样{    int i,j;    next[1]=0;    i=1;j=1; //这里ij分别是二者的起始位置    while(i<=n&&j<=m)    {        if(j==0||a[i]==b[j]) {i++;j++;}//如果相等则比较下一个  不等就回溯 这里不要再给next数组赋值了上面已经赋好了        else            j=next[j];    }    if(j>m) return i-m;    else return -1;}int main(){      int i,t,ans;      scanf("%d",&t);      while(t--)      {          scanf("%d %d",&n,&m);          for(i=1;i<=n;i++)              scanf("%d",&a[i]);//注意从1开始输入          for(i=1;i<=m;i++)              scanf("%d",&b[i]);          get_next();          ans=KMP();          printf("%d\n",ans);      }      return 0;}

对next 数组现在已有一个基本的了解：例如当文本串s[3]!=p[3]时，i=3不动，j=next[j]=next[3],j就有一个新的值

进行比较了。。。。。我理解的next 数组就这么简单

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