HDU 1711 -Number Sequence(KMP)
题目
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 40385 Accepted Submission(s): 16656
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
代码
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=1e6 + 10;
int a[MAXN],b[MAXN];
int nxt[MAXN];
int n,m;
void getnxt()
{
int i=1;
int j=0;
nxt[0]=0;
while(i<m)
{
if(b[i]==b[j])
{
nxt[i++]=++j;
}
else if(!j)
{
i++;
}
else
{
j=nxt[j-1];
}
}
}
int kmp()
{
int i=0;
int j=0;
while(i<n && j<m)
{
if(a[i]==b[j])
{
i++;
j++;
}
else if(!j)
{
i++;
}
else
{
j=nxt[j-1];
}
}
if(j==m) return i-m+1;
else return -1;
}
int main()
{
int t,i,k;
scanf("%d",&t);
while(t--)
{
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]);
getnxt();
cout<<kmp()<<endl;
}
return 0;
}
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，太厉害了
