题目链接:https://cn.vjudge.net/contest/248066#problem/A
Number Sequence
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
KMP 算法
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 1e6+100;
int nxt[MAXN];
int s1[MAXN];
int s2[MAXN];
int n,m;
void getnxt(){
int i=1;
int j=0;
nxt[0]=0;
while(i<n)
{
if(s1[i]==s1[j]){
nxt[i]=j+1;
i++;
j++;
}else if(!j){
i++;
}else{
j=nxt[j-1];
}
}
}
int kmp(){
int i=0;
int j=0;
while(j<n&& i<m){
if(s1[j]==s2[i]){
j++;
i++;
}else if(!j){
i++;
}else{
j=nxt[j-1];
}
if(j==n){
return i-n+1;
}
}
return -1;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&m,&n);
for(int i=0;i<m;i++){
scanf("%d",&s2[i]);
}
for(int i=0;i<n;i++){
scanf("%d",&s1[i]);
}
getnxt();
cout<<kmp()<<endl;
}
return 0;
}