Codeforces Round #600 (Div. 2) - B. Silly Mistake(模拟)

The meaning of problems: a company with employees in and out every day, $ a [i]> represents $ a [i] 0 $ $ when the employee into the company, $ a [i] <represents $ -a [i] 0 $ when $ the staff of the company, there are some strict rules and out of office

• Office staff can only enter once a day
• If he did that day into the office, he obviously can not leave
• Start and end of the office are empty (employees can not stay the night) a day, the office may at any time of the day are empty

You are given a sequence $ a $ ($ a [i] \ neq 0 $), ask if you can put an array $ a $ is divided into several adjacent sub-arrays, each sub-array out of the situation so that employees comply with the requirements and the length of each sub-array output, or the output can not represent $ $ -1

Thinking: MK $ $ array with a flag state of a person, $ mk [i] = 0 $ $ I $ denotes the number of employees within has not been operated, $ mk [i] = 1 $ $ I $ denotes the number of employees the days have been into the office, $ mk [i] = - 1 $ $ i $ represents the number of days that employees have entered the office and out, $ num $ record number of office at this time, if when $ num = $ 0 you can enter a new day, while the need to use an array of $ p $ records for this day has to go out of employee numbers, when entering a new day, you need these employees $ mk [i] $ becomes $ 0 $, appeared in the middle not lawful directly out of the loop, the output of $ -1 $

 

#include <iostream>
#include <algorithm>
#include <cstdio>

using namespace std;

const int N = 100010;
const int M = 1000010;

int n, a[N], mk[M];
int num, cnt, p[N];
int res[N], pos;

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int flag = 1; res[++pos] = 0;
    for (int i = 1; i <= n; i++) {
        if (a[i] > 0) {
            if (0 == mk[a[i]]) {
                mk[a[i]] = 1, num++;
            }
            else {
                flag = 0; break;
            }
        }
        else {
            if (1 == mk[-a[i]]) {
                mk[-a[i]] = -1, num--;
                p[++cnt] = -a[i];
                if (0 == num) {
                    for (int j = 1; j <= cnt; j++) mk[p[j]] = 0;
                    cnt = 0, res[++pos] = i;
                }
            }
            else {
                flag = 0; break;
            }
        }
    }
    if (0 != num) flag = 0;
    if (!flag) printf("-1\n");
    else {
        printf("%d\n", pos - 1);
        for (int i = 2; i <= pos; i++) {
            printf("%d", res[i] - res[i - 1]);
            printf(i == pos ? "\n" : " ");
        }
    }
    return 0;
}