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

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：

 #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 ;
 }