HDU 1711 Number Sequence KMP应用 求成功匹配子串的最小下标

Posted 2021-01-22 ymzjj

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42917    Accepted Submission(s): 17715

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

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

 

题意概括：

给出一串主串，一串子串；

求成功匹配到子串的最小下标；

 

解题思路：

KMP的应用，稍微变形，匹配到子串就跳出来输出。

 

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #define INF 0x3f3f3f3f
 7 using namespace std;
 8 const int MAXN = 1e6+10;
 9 const int MAXM = 1e4+10;
10 int W[MAXM], T[MAXN];
11 int wlen, tlen;
12 int nxt[MAXM];
13 
14 void get_nxt()
15 {
16     int j, k;
17     j = 0;
18     k = -1;
19     nxt[0] = -1;
20     while(j < wlen){
21         if(k == -1 || W[j] == W[k]){
22             nxt[++j] = ++k;
23         }
24         else k = nxt[k];
25     }
26 }
27 
28 int KMP_index()
29 {
30     int i = 0, j = 0;
31     get_nxt();
32 
33     while( i < tlen && j < wlen){
34         if(j == -1 || T[i] == W[j]){
35             i++;
36             j++;
37         }
38         else j = nxt[j];
39     }
40     if(j == wlen) return i-wlen+1;
41     else return -1;
42 }
43 
44 int main()
45 {
46     int T_case, N, M;
47     scanf("%d", &T_case);
48     while(T_case--)
49     {
50         scanf("%d%d", &N, &M);
51         wlen = M, tlen = N;
52         for(int i = 0; i < N; i++)
53             scanf("%d", &T[i]);
54         for(int j = 0; j < M; j++)
55             scanf("%d", &W[j]);
56 
57         printf("%d
", KMP_index());
58     }
59     return 0;
60 }

View Code