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
第一道kmp 是一道模板题
不过还是wa了两发
1.不能开next数组 和系统自带名字重名
2.一开始我是把一个个数据转化成一条字符串来进行kmp 但是显然是错的QWQ
改成数组形式就行了
#include<bits/stdc++.h>
using namespace std;
//input
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);i--)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m);
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define inf 0x3f3f3f3f
#define REP(i,N) for(int i=0;i<(N);i++)
#define CLR(A,v) memset(A,v,sizeof A)
//////////////////////////////////
int nex[100000+5];
int lenp,lens;
int p[10000+5];//可以是char 也可以是string
int s[1000000+5];
void getnext()
{
nex[0]=-1;
int k=-1,j=0;
while(j<lenp-1)
{
if(k==-1||p[j]==p[k])
nex[++j]=++k;
else k=nex[k];
}
}
int kmp()
{
//lens=
//lenp=
int j=0;
int i=0;
while(i<lens&&j<lenp)
{
if(s[i]==p[j]||j==-1)
{
i++;
j++;
}
else
j=nex[j];
if(j==lenp)
{
return i-j+1;
}
}
return -1;
}
int main()
{
int cas;
RI(cas);
while(cas--)
{
RII(lens,lenp);
rep(i,0,lens-1)
RI(s[i]);
rep(i,0,lenp-1)
RI(p[i]);
getnext();
cout<<kmp()<<endl;
}
return 0;
}
View Code
Number Sequence kmp
原文：https://www.cnblogs.com/bxd123/p/10672931.html
