Educational Codeforces Round 152 (Rated for Div. 2)

D. Array Painting

https://codeforces.com/contest/1849/problem/D

2 seconds / 256 megabytes

standard input / standard output

## Problem

You are given an array of $n$ integers, where each integer is either $0$, $1$, or $2$. Initially, each element of the array is blue.

Your goal is to paint each element of the array red. In order to do so, you can perform operations of two types:

• pay one coin to choose a blue element and paint it red;
• choose a red element which is not equal to $0$ and a blue element adjacent to it, decrease the chosen red element by $1$, and paint the chosen blue element red.
What is the minimum number of coins you have to spend to achieve your goal?

## Input

The first line contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 2$).

## Output

Print one integer — the minimum number of coins you have to spend in order to paint all elements red.

Input

3
0 2 0

Output

1

Input

4
0 0 1 1

Output

2

Input

7
0 1 0 0 1 0 2

Output

4

## Note

In the first example, you can paint all elements red with having to spend only one coin as follows:

1. paint the $2$-nd element red by spending one coin;
2. decrease the $2$-nd element by $1$ and paint the $1$-st element red;
3. decrease the $2$-nd element by $1$ and paint the $3$-rd element red.

In the second example, you can paint all elements red spending only two coins as follows:

1. paint the $4$-th element red by spending one coin;
2. decrease the $4$-th element by $1$ and paint the $3$-rd element red;
3. paint the $1$-st element red by spending one coin;
4. decrease the $3$-rd element by $1$ and paint the $2$-nd element red.

## 题解

• 第 $i$ 位为 $0$：花一个硬币“激活”为后，只有第 $i$ 位变红，总区间为 $[i,i]$。
• 第 $i$ 位为 $1$：花一个硬币“激活”为后，除了第 $i$ 位变红还有：

• 如果选择左侧，则从第 $i-1$ 位开始到左侧第一个 $0$ 结束都能链式变红，总区间为 $[左侧第一个0,i]$.
• 如果选择右侧，则从第 $i+1$ 位开始到右侧第一个 $0$ 结束都能链式变红，总区间为 $[i,右侧第一个0]$.
• 第 $i$ 位为 $2$：花一个硬币“激活”为后，除了第 $i$ 位变红还有：

• 对于左侧，从第 $i-1$ 位开始到左侧第一个 $0$ 结束都能链式变红。
• 对于右侧，从第 $i+1$ 位开始到右侧第一个 $0$ 结束都能链式变红。
• 总区间为 $[左侧第一个0,右侧第一个0]$

## 代码

#include <bits/stdc++.h>
#define endl '\n'
#define int long long

using namespace std;

void solve()
{
int n;
cin >> n;
vector<int> a(n + 10);
for (int i = 1; i <= n; i++)
cin >> a[i];
vector<int> pre(n + 10), nxt(n + 10);
for (int i = 1; i <= n; i++)
{
if (a[i - 1])
pre[i] = pre[i - 1];
else
pre[i] = i - 1;
}
for (int i = n; i >= 1; i--)
{
if (a[i + 1])
nxt[i] = nxt[i + 1];
else
nxt[i] = i + 1;
}
// for (int i = 1; i <= n; i++)
//     cout << pre[i] << " \n"[i == n];
// for (int i = 1; i <= n; i++)
//     cout << nxt[i] << " \n"[i == n];
vector<pair<int, int>> pv;
for (int i = 1; i <= n; i++)
{
if (a[i] == 0)
{
pv.push_back({i, i});
}
else if (a[i] == 1)
{
pv.push_back({i, nxt[i]});
pv.push_back({pre[i], i});
}
else // if (a[i] == 2)
{
pv.push_back({pre[i], nxt[i]});
}
}
sort(pv.begin(), pv.end());
int ans = 0, idx = 0;
for (int i = 1; i <= n; i++)
{
int l = -1, r = -1;
for (; idx < pv.size(); idx++)
{
if (!(l == -1 || pv[idx].first <= i))
break;
l = pv[idx].first;
r = pv[idx].second;
}
ans += max(0LL, l - i);
ans++;
i = r;
}
cout << ans << endl;
}

signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
// cin >> t;
while (t--)
solve();
return 0;
}