Codeforces Round #215 (Div. 1) B. Sereja ans Anagrams 匹配

B. Sereja ans Anagrams

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/367/B

Description

Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b consists of m integers b1, b2, ..., bm. As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q (q + (m - 1)·p ≤ n; q ≥ 1), such that sequence b can be obtained from sequence aq, aq + p, aq + 2p, ..., aq + (m - 1)p by rearranging elements.

Sereja needs to rush to the gym, so he asked to find all the described positions of q.

[b]Input[/b]

The first line of the input data contains the single integer n (1 ≤ n ≤ 50) — the number of commands.

Then
follow n lines, each contains one command. Each of these lines contains
either command pwd, or command cd, followed by a space-separated
non-empty parameter.

The command parameter cd only contains lower
case Latin letters, slashes and dots, two slashes cannot go
consecutively, dots occur only as the name of a parent pseudo-directory.
The command parameter cd does not end with a slash, except when it is
the only symbol that points to the root directory. The command parameter
has a length from 1 to 200 characters, inclusive.

Directories in the file system can have the same names.

[b]Output[/b]

The first line contains three integers n, m and p (1 ≤ n, m ≤ 2·105, 1 ≤ p ≤ 2·105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109).

[b]Sample Input[/b]

5 3 1
1 2 3 2 1
1 2 3

[b]Sample Output[/b]

2
1 3

HINT

[b]题意
[/b]

给你n个a[i]，m个b[i]，和P

要求让你找到起点，使得a[pos],a[pos+p],a[pos+2*p]……,a[pos+(m-1)*p]和b数组一样

[b]题解：[/b]

类似于km算法

我们把b[i]表示为子串

a[i]表示为母串

因为不要求顺序，我们就直接跑就好了，只要匹配不是O(m*n)的都能过吧

[b]代码[/b]

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 1050005
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
//**************************************************************************************

int a[maxn];
int b[maxn];
map<int,int> H;
map<int,int> H1;
vector<int> ans;
int main()
{
int n=read(),m=read(),p=read();
for(int i=1;i<=n;i++)
a[i]=read();
for(int i=1;i<=m;i++)
b[i]=read(),H[b[i]]++;
int pos;
int flag=0;
for(int i=1;i<=p;i++)
{
H1.clear(),pos=i,flag=0;;
for(int j=i;j<=n;j+=p)
{
H1[a[j]]++,flag++;;
if(H1[a[j]]>H[a[j]])
{
for(int k=pos;k<=j;k+=p)
{
H1[a[k]]--,flag--;
if(a[j]==a[k])
{
pos=k+p;
break;
}
}
continue;
}
if(flag==m)
{
ans.push_back(pos);
H1[a[pos]]--;
flag--;
pos+=p;
}
}
}
sort(ans.begin(),ans.end());
cout<<ans.size()<<endl;
for(int i=0;i<ans.size();i++)
printf("%d ",ans[i]);

}