這個題有點意思，一開始沒想到用dp，沒啥思路，後來看題解才恍然大悟：k才1~100，直接枚舉每個-1點的k取值進行dp就行了。先預處理出來sz[i][j] i左邊的比j大的數，lz[i][j] i右邊比j小的數。然後就沒啥了。

題幹：

Description

Input

Output

Sample Input

Sample Output

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include
 
<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(int i = a;i <= n;i++)
#define lv(i,a,n) for(int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template 
 
<class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < ‘0‘ || c > ‘9‘)
        if(c == ‘-‘) op = 1;
    x = c - ‘0‘;
    while(c = getchar(), c >= ‘0‘ && c <= ‘9‘)
        x = x * 10 + c - ‘0‘;
    if(op) x = -x;
}
template <class
 
 T>
void write(T x)
{
    if(x < 0) putchar(‘-‘), x = -x;
    if(x >= 10) write(x / 10);
    putchar(‘0‘ + x % 10);
}
int n,m,tot = 0;
int a[10020],num[10020];
int lz[10050][105];
int sz[10020][105];
int dp[10020][105];
int k;
int main()
{
    read(n);read(k);
    duke(i,1,n)
    {
        read(a[i]);
        if(a[i] == -1)
        num[++tot] = i;
    }
    duke(i,1,n) 
    {
        duke(j,1,k)
        {
            if(a[i - 1] > j)
                sz[i][j] = sz[i - 1][j] + 1;
            else
                sz[i][j] = sz[i - 1][j];
        }
    }
    lv(i,n - 1,1)
    {
        duke(j,1,k)
        {
            if(a[i + 1] < j && a[i + 1] != -1)
                lz[i][j] = lz[i + 1][j] + 1;
            else
                lz[i][j] = lz[i + 1][j];
        }
    }
    duke(i,1,n)
    {
        duke(j,1,k)
        {
            if(a[i] != -1)
            {
                dp[i][j] = dp[i - 1][j];
                continue;
            }
            else
            {
                dp[i][j] = INF;
                duke(l,1,k)
                {
                    dp[i][j] = min(dp[i - 1][l] + lz[i][j] + sz[i][j],dp[i][j]);
                }
            }
        }
    }
    int ans = INF;
    duke(i,1,k)
    {
        ans = min(ans,dp[n][i]);
    }
    duke(i,1,n)
    {
        if(a[i] != -1)
        {
            ans += sz[i][a[i]];
        }
    }
    printf("%d\n",ans);
    return 0;
}
/*
5 4
4 2 -1 -1 3
*/

B1786 [Ahoi2008]Pair 配對 逆序對+dp