Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10571 Accepted Submission(s): 4814
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
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn1=1000010;
const int maxn2=10010;
int n,m;
int a[maxn1];
int b[maxn2];
int next[maxn2];
int main()
{
int kmp (int *a,int *b,int *next);
int t,i;
cin>>t;
while(t--)
{
cin>>n>>m;
for(i=0;i<n;i++)
{scanf("%d",&a[i]);}
for(i=0;i<m;i++)
{scanf("%d",&b[i]);}
int ans=kmp(a,b,next);
cout<<ans<<endl;
}
return 0;
}
//此函数用来求匹配串s串的next数组
void getnext (int *s,int *next)
{
next[0]=next[1]=0;
for (int i=1;i<m;i++)//m为匹配串s的长度
{
int j=next[i];
while (j&&s[i]!=s[j])
j=next[j];
next[i+1]=s[i]==s[j]?j+1:0;
}
}
//此函数用来求匹配位置的值的函数，匹配不成功返回值-1
int kmp (int *a,int *b,int *next)
{
getnext (b,next);/////////
int j=0;
for (int i=0;i<n;i++)
{/////n为串1的长度
while (j&&a[i]!=b[j])
j=next[j];
if (a[i]==b[j])
j++;
if (j==m)//m为串2的长度
return i-m+2;
}
return -1;
}
HDU 1711 Number Sequence(字符串匹配)