kuangbin专题十六 KMP&&扩展KMP HDU1711 Number Sequence

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.

InputThe 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].

OutputFor 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

首先是预处理Next数组，初始化Next[0]=-1,j=-1  然后从i=0开始，如果j!=-1&&不匹配，j=Next[j]。否则判断p[++i]==p[++j],如果相等，直接让Next[i]=Next[j]，不等的话，让Next[i]=j;

然后在主串里匹配。i=0,j=0; j!=-1&&不匹配， j=Next[j]。否则i++，j++

 #include<stdio.h>

 #include<string.h>

 int Next[],t[],p[];//模式串对应的Next

 int n,m,_;

 void prekmp() {//预处理Next数组

     int i,j;

     j=Next[]=-;

     i=;

     while(i<m) {

         while(j!=-&&p[i]!=p[j]) j=Next[j];

         if(p[++i]==p[++j]) Next[i]=Next[j];//会快一点 博客体现了这一点

         else Next[i]=j;

     }

 }

 int kmp() {//查找在主串的位置

     int i=,j=;

     prekmp();

     while(i<n&&j<m) {

         while(j!=-&&t[i]!=p[j]) j=Next[j];

         i++;j++;

     }

     if(j==m) return i-m+;

     else return -;

 }

 int main() {

     for(scanf("%d",&_);_;_--) {

         scanf("%d%d",&n,&m);

         for(int i=;i<n;i++) {

             scanf("%d",&t[i]);

         }

         for(int i=;i<m;i++) {

             scanf("%d",&p[i]);

         }

         int ans=kmp();

         printf("%d\n",ans);

     }

 }

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kuangbin专题十六 KMP&&扩展KMP HDU1711 Number Sequence

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