题目描述:
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.
输入：
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].
输出：
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
样例输入：
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
样例输出：
6
-1
KMP模板题。
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<malloc.h>
#include<math.h>
#include<string>
#include<queue>
#include<stack>
#include<cstring>
#include<vector>
#define inf 0x3f3f3f3f
#define MAXN 1000000+10
#define LL long long
#define MAX 1000000007
#define pi 3.1415926
using namespace std;
int a[MAXN],b[MAXN],next1[MAXN];
int n,m;
void next_cal(int len)
{
next1[0]=-1;
int k=-1;
for(int q=1;q<=len-1;q++)
{
while(k>-1&&b[k+1]!=b[q])
k=next1[k];
if(b[k+1]==b[q])
k+=1;
next1[q]=k;
}
}
int KMP()
{
next_cal(m);
int k=-1;
for(int i=0;i<n;i++)
{
while(k>-1&&b[k+1]!=a[i])
{
k=next1[k];
}
if(b[k+1]==a[i])
k+=1;
if(k==m-1)
return i-m+1;
}
return -1;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(next1,-1,sizeof(next1));
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
int ans=KMP();
if(ans!=-1)
printf("%d\n",ans+1);
else
printf("-1\n");
}
return 0;
}