Codeforces 467e Alex and complicated task (efficient)

Link: codeforces 467e Alex and complicated task

Given a sequence of N lengths, select as many 4 tuples as possible (not overlapping ).

Solution: The left end of every four tuples must be as small as possible. Therefore, a monotonous stack is used for maintenance. If the newly added number x appears in the stack, mark the number between two numbers as X. If the tag of a number is not empty, it means that the corresponding four tuples are found. Because the sequence is traversed from the left, it must be the best.

#include <cstdio>#include <cstring>#include <map>#include <stack>#include <vector>#include <algorithm>using namespace std;const int maxn = 5 * 1e5 + 5;int N, B[maxn];void solve () {    vector<int> ans;    map<int, int> pre, sum;    int mv = 0;    while (mv < N) {        pre.clear();        sum.clear();        stack<int> sta;        for (int& i = mv; i < N; i++) {            if (pre[B[i]]) {                for (int j = 0; j < 2; j++) {                    ans.push_back(pre[B[i]]);                    ans.push_back(B[i]);                }                break;            }            while (!sta.empty() && (sum[B[i]] > 1 || (sum[B[i]] == 1 && sta.top() != B[i]))) {                pre[sta.top()] = B[i];                sum[sta.top()]--;                sta.pop();            }            sta.push(B[i]);            sum[B[i]]++;        }        mv++;    }    printf("%lu\n", ans.size());    for (int i = 0; i < ans.size(); i++)        printf("%d%c", ans[i], i == ans.size() - 1 ? ‘\n‘ : ‘ ‘);}int main () {    scanf("%d", &N);    for (int i = 0; i < N; i++)        scanf("%d", &B[i]);    solve();    return 0;}

Codeforces 467e Alex and complicated task (efficient)