Kobolds and Catacombs 思维，模拟，前缀，后缀（沈阳）

题意 ：

• 给一序列，将其分为若干段，要求每一小段内排序后整体非递减，求最多能分为多少段

思路 ：

• 对一段内部，必然有最大值和最小值，要让分的段数最多，只要让下一段的最小值大于这段的最大值，就可以多分一段
• 所以直接找每个数能接受的最大值和最小值
• 即找这个数往前的最大值，下个数往后的最小值（找时注意设立两个哨兵）

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
using namespace std;
typedef long long ll;

const int N = 1e6 + 10;

int n;
int a[N];
int maxx[N], minn[N];

int main()
{
    
    
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    
    int _ = 1;
//    cin >> _;

    while (_ -- )
    {
    
    
        cin >> n;
        for (int i = 1; i <= n; i ++ ) cin >> a[i];
        
        maxx[0] = 0;
        minn[n + 1] = 1e9 + 7;
        
        for (int i = 1; i <= n; i ++ )
            maxx[i] = max(maxx[i - 1], a[i]);
        for (int i = n; i >= 1; i -- )
            minn[i] = min(minn[i + 1], a[i]);
        
        int ans = 0;
        for (int i = 1; i <= n; i ++ )
            if (maxx[i] <= minn[i + 1])
                ans ++ ;
        
        cout << ans << endl;
    }
    
    return 0;
}

• 附上另一种方案，只记录后缀的，且不增加哨兵

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
using namespace std;
typedef long long ll;

const int N = 1e6 + 10;

int n;
int a[N];
int maxx[N], minn[N];

int main()
{
    
    
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    
    int _ = 1;
//    cin >> _;

    while (_ -- )
    {
    
    
        cin >> n;
        for (int i = 1; i <= n; i ++ ) cin >> a[i];
        
        int mi = a[n];
        for (int i = n; i >= 1; i -- )
        {
    
    
            mi = min(mi, a[i]);
            minn[i] = mi;
        }
        
        int mx = 0, ans = 0;
        for (int i = 1; i <= n; i ++ )
        {
    
    
            if (mx <= minn[i]) ans ++ ;
            mx = max(mx, a[i]);
        }
        
        cout << ans << endl;
    }
    
    return 0;
}