Codeforces 367B Sereja ans Anagrams 【map维护queue】

题目链接：Codeforces 367B Sereja ans Anagrams

B. Sereja ans Anagrams
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Input
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).

Output
In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.

Examples
input
5 3 1
1 2 3 2 1
1 2 3
output
2
1 3
input
6 3 2
1 3 2 2 3 1
1 2 3
output
2
1 2

题意：给定n个元素的序列a[]和m个元素的序列b[]，让你找到一个q使得a[q]、a[q+p]、…a[q+(m-1)*p]构成序列b。问你所有的q。

思路：i、i+p、i+2*p肯定构成一条长链，枚举链的起点，用一个queue来维护元素的进出，每当queue里面有m个元素，用map判定是否满足。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#define PI acos(-1.0)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 2*1e5 + 10;
const int pN = 1e6;// <= 10^7
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
int a[MAXN];
map<int, int> fp, tp;
int n, m, p;
int ans;
bool vis[MAXN];
void Solve(int s) {
    tp.clear(); queue<int> Q;
    for(int i = s; i <= n; i += p) {
        Q.push(i); tp[a[i]]++;
        if(Q.size() == m) {
            if(fp == tp) {
                vis[Q.front()] = true;
                ans++;
            }
            int v = a[Q.front()]; Q.pop();
            if(--tp[v] == 0) {
                tp.erase(v);
            }
        }
    }
}
int main()
{
    scanf("%d%d%d", &n, &m, &p);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        vis[i] = false;
    }
    fp.clear();
    for(int i = 1; i <= m; i++) {
        int b; scanf("%d", &b);
        fp[b]++;
    }
    ans = 0;
    for(int i = 1; i <= p; i++) {
        Solve(i);
    }
    printf("%d\n", ans); int cnt = 0;
    for(int i = 1; i <= n; i++) {
        if(vis[i]) {
            if(cnt > 0) printf(" ");
            printf("%d", i); cnt++;
        }
    }
    if(ans) printf("\n");
    return 0;
}