字符串_KMP算法(求next[]模板 hdu 1711)

浏览数：17 / 时间：2015年06月08日

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

问题描述：给两个序列a,b，长度分别为n,m(1<=n<=1000000,1<=m<=10000),问序列b是否为序列a的子序列，若是：返回a中最左边的与b相等的子序列的首元素下标；若不是，输出-1。

目的：方便以后查看KMP算法中next[]的模板

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12811 Accepted Submission(s): 5815

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

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

-1

代码实现：

 1 #include "stdio.h"
 2 #include "string.h"
 3 #define N 10005
 4 
 5 int next[N];
 6 int a[100*N],b[N];
 7 
 8 void KMP(int *s,int len,int *next) //求next[]模板
 9 {
10     int i,j;
11     i = 0;
12     j = next[0] = -1;
13     while(i<len)
14     {
15         while(j!=-1 && s[i]!=s[j])
16             j = next[j];
17         next[++i] = ++j;
18     }
19 }
20 
21 int main()
22 {
23     int T;
24     int i,j;
25     int n,m;
26     scanf("%d",&T);
27     while(T--)
28     {
29         scanf("%d %d",&n,&m);
30         for(i=0; i<n; i++) scanf("%d",&a[i]);
31         for(i=0; i<m; i++) scanf("%d",&b[i]);
32         KMP(b,m,next);
33         i=j=0;
34         while(i<n)
35         {
36             while(j!=-1 && a[i]!=b[j])
37                 j = next[j];
38             i++,j++;
39             if(j==m) break;
40         }
41         if(j==m) printf("%d\n",i-j+1); //题中数据是从下标为1开始的，故加1
42         else printf("-1\n");
43     }
44     return 0;
45 }