题意描述
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.
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
思路
不难看出这是一道kmp板子题，直接套模板即可。对于无法匹配的情况，我们可以使用一个标记来记录，需要注意的一点是，输入的值可能会有负数，所以我们要使用int数组来存储a和b两个序列
AC代码
代码语言：javascript复制#include<iostream>
#include<algorithm>
#include<cstring>
#define IOS ios::sync_with_stdio(false)
using namespace std;
const int maxn=1e6 10;
const int maxm=10010;
int w[maxm],t[maxn];
int ne[maxm];
int n;
int main(){
IOS;cin.tie(0);
cin>>n;
while(n--){
int x,y;
bool flag=true;
cin>>x>>y;
for(int i=1;i<=x;i ) cin>>t[i];
for(int i=1;i<=y;i ) cin>>w[i];
for(int i=2,j=0;i<=y;i ){
while(j&&w[i]!=w[j 1]) j=ne[j];
if(w[i]==w[j 1]) j ;
ne[i]=j;
}
for(int i=1,j=0;i<=x;i ){
while(j&&t[i]!=w[j 1]) j=ne[j];
if(t[i]==w[j 1]) j ;
if(j==y){
flag=false;
cout<<i-y 1<<endl;
break;
}
}
if(flag) cout<<"-1"<<endl;
}
}
