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

时间：2024-11-09 11:38:30

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

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：

 #include <cstdio>

 #include <iostream>

 #include <cstring>

 #include <algorithm>

 #include <cmath>

 #define INF 0x3f3f3f3f

 using namespace std;

 const int MAXN = 1e6+;

 const int MAXM = 1e4+;

 int W[MAXM], T[MAXN];

 int wlen, tlen;

 int nxt[MAXM];

 void get_nxt()

 {

     int j, k;

     j = ;

     k = -;

     nxt[] = -;

     while(j < wlen){

         if(k == - || W[j] == W[k]){

             nxt[++j] = ++k;

         }

         else k = nxt[k];

     }

 }

 int KMP_index()

 {

     int i = , j = ;

     get_nxt();

     while( i < tlen && j < wlen){

         if(j == - || T[i] == W[j]){

             i++;

             j++;

         }

         else j = nxt[j];

     }

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

     else return -;

 }

 int main()

 {

     int T_case, N, M;

     scanf("%d", &T_case);

     while(T_case--)

     {

         scanf("%d%d", &N, &M);

         wlen = M, tlen = N;

         for(int i = ; i < N; i++)

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

         for(int j = ; j < M; j++)

             scanf("%d", &W[j]);

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

     }

     return ;

 }

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