HDU - 1711 Number Sequence

Time Limit: 5000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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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






裸kmp,可作为模板:
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1000000+10;
const int maxm=10000+10;
int next[maxm],p[maxm],s[maxn];
int n,m;
void getnext(int *p)
{
    int j=-1,i=0;
    next[0]=-1;
    while(i<m-1)
    {
        if(j==-1||p[i]==p[j])
        {
            i++,j++;
            next[i]=j;
        }
        //else j=next[j];
        else j=-1;//还是这句简洁高效
    }
}


int kmp(int *p,int *s)
{
    int j=0,i=0;
    while(i<n&&j<m)
    {
        if(j==-1||s[i]==p[j])
        {
            i++,j++;
        }
        else j=next[j];
    }
    if(j==m) return i-j+1;
    else return -1;
}


int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
      scanf("%d%d",&n,&m);
      for(i=0;i<n;i++)
      scanf("%d",&s[i]);
      for(j=0;j<m;j++)
      scanf("%d",&p[j]);
      getnext(p);
      printf("%d\n",kmp(p,s));
    }
}


这里讲kmp原理,还不错:
http://kb.cnblogs.com/page/176818/
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