hdu 1711Number Sequence (KMP入门,子串第一次出现的位置)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15149    Accepted Submission(s): 6644

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：给两个串a,b;求b在a中首次出现的位置，若没有則输出-1;

 #include<iostream>
 #include<vector>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <math.h>
 #include<algorithm>
 #define ll long long
 #define eps 1e-8
 using namespace std;

 int nexts[];
 int  str[],b[];

 void pre_nexts(int m)
 {
     memset(nexts,,sizeof(nexts));
     int j = ,k = -;
     nexts[] = -;
     while(j < m)
     {
         if(k == - || b[j] == b[k])  nexts[++j] = ++k;
         else k = nexts[k];
     }
 }
 int KMP(int n,int m)
 {
     int i,j = ;
     for(i = ; i < n; )
     {
         if(str[i] == b[j])
         {
             if(j == m-) return i - (m - )+;//匹配成功，返回起始坐标
             i++,j++;
         }
         else
         {
             j = nexts[j];
             if(j == -) { i++,j = ;}//否则重新匹配
         }
     }
     return -;
 }
 int main(void)
 {
     int t,i,m,n;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d %d",&n,&m);
         for(i = ; i < n; i++)
             scanf("%d",&str[i]);
         for(i = ; i < m; i++)
             scanf("%d",&b[i]);
         if(n < m) { printf("-1\n"); continue;}//细节
         pre_nexts(m);
         printf("%d\n",KMP(n,m));
     }
     return ;
 }