B1786 [Ahoi2008]Pair 配对 逆序对+dp

这个题有点意思，一开始没想到用dp，没啥思路，后来看题解才恍然大悟：k才1~100，直接枚举每个-1点的k取值进行dp就行了。先预处理出来sz[i][j]  i左边的比j大的数，lz[i][j]  i右边比j小的数。然后就没啥了。

题干：

Description

Input

Output

Sample Input

5 4
4 2 -1 -1 3

Sample Output

4

代码：

#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 =  << ;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = ;
    while(c = getchar(), c < '' || c > '')
        if(c == '-') op = ;
    x = c - '';
    while(c = getchar(), c >= '' && c <= '')
        x = x *  + c - '';
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < ) putchar('-'), x = -x;
    if(x >= ) write(x / );
    putchar('' + x % );
}
int n,m,tot = ;
int a[],num[];
int lz[][];
int sz[][];
int dp[][];
int k;
int main()
{
    read(n);read(k);
    duke(i,,n)
    {
        read(a[i]);
        if(a[i] == -)
        num[++tot] = i;
    }
    duke(i,,n)
    {
        duke(j,,k)
        {
            if(a[i - ] > j)
                sz[i][j] = sz[i - ][j] + ;
            else
                sz[i][j] = sz[i - ][j];
        }
    }
    lv(i,n - ,)
    {
        duke(j,,k)
        {
            if(a[i + ] < j && a[i + ] != -)
                lz[i][j] = lz[i + ][j] + ;
            else
                lz[i][j] = lz[i + ][j];
        }
    }
    duke(i,,n)
    {
        duke(j,,k)
        {
            if(a[i] != -)
            {
                dp[i][j] = dp[i - ][j];
                continue;
            }
            else
            {
                dp[i][j] = INF;
                duke(l,,k)
                {
                    dp[i][j] = min(dp[i - ][l] + lz[i][j] + sz[i][j],dp[i][j]);
                }
            }
        }
    }
    int ans = INF;
    duke(i,,k)
    {
        ans = min(ans,dp[n][i]);
    }
    duke(i,,n)
    {
        if(a[i] != -)
        {
            ans += sz[i][a[i]];
        }
    }
    printf("%d\n",ans);
    return ;
}
/*
5 4
4 2 -1 -1 3
*/