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杭电 1711 Number Sequence
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10239 Accepted Submission(s): 4656
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
经典 kmp。
AC代码例如以下：
#include<iostream>
#include<cstdio>
using namespace std;
int a[1000005],b[10005],next[10005];
int main()
{
int t;
int n,m,ans;
int i,j;
scanf("%d",&t);
while(t--)
{
ans=-1;
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<m;i++)
scanf("%d",&b[i]);
i=0;j=-1;next[0]=-1;
while(i<m)
{
if(j==-1||b[i]==b[j])
next[++i]=++j;
else
j=next[j];
}
i=0;j=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j])
i++,j++;
else j=next[j];
if(j==m)
{
ans=i-j+1;break;
}
}
printf("%d\n",ans);
}
return 0;
}
杭电 1711 Number Sequence
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