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++


 1 #include<stdio.h>
 2 #include<string.h>
 3 int Next[10010],t[1000010],p[10010];//模式串对应的Next
 4 int n,m,_;
 5 
 6 void prekmp() {//预处理Next数组
 7     int i,j;
 8     j=Next[0]=-1;
 9     i=0;
10     while(i<m) {
11         while(j!=-1&&p[i]!=p[j]) j=Next[j];
12         if(p[++i]==p[++j]) Next[i]=Next[j];//会快一点 博客体现了这一点
13         else Next[i]=j;
14     }
15 }
16 
17 int kmp() {//查找在主串的位置
18     int i=0,j=0;
19     prekmp();
20     while(i<n&&j<m) {
21         while(j!=-1&&t[i]!=p[j]) j=Next[j];
22         i++;j++;
23     }
24     if(j==m) return i-m+1;
25     else return -1;
26 }
27 
28 int main() {
29     for(scanf("%d",&_);_;_--) {
30         scanf("%d%d",&n,&m);
31         for(int i=0;i<n;i++) {
32             scanf("%d",&t[i]);
33         }
34         for(int i=0;i<m;i++) {
35             scanf("%d",&p[i]);
36         }
37         int ans=kmp();
38         printf("%d\n",ans);
39     }
40 }

 






posted @ 2019-01-13 17:39  Frontierone  阅读(130)  评论(0编辑  收藏  举报