luoguP4005 小 Y 和地铁

luoguP4005 小 Y 和地铁

题目描述

链接

Solution

#include<bits/stdc++.h>

using namespace std;

inline int read()
{
    int f = 1,x = 0;
    char ch;
    do
    {
        ch = getchar();
        if(ch == '-')f = -1;
    }while(ch<'0'||ch>'9');
    do
    {
        x = (x<<3) + (x<<1) + ch - '0';
        ch = getchar();
    }while(ch>='0'&&ch<='9');
    return f*x;
}

const int MAXN = 44 + 10;

int T;
int n;
int a[MAXN];
int l[MAXN],r[MAXN],num;
int ans;

int c_up[MAXN],c_down[MAXN];

inline int lowbit(int x)
{
    return x&(-x);
}

inline void add1(int x,int t)
{
    while(x<=n)
    {
        c_up[x] += t;
        x+=lowbit(x);
    }
}

inline void add2(int x,int t)
{
    while(x<=n)
    {
        c_down[x] += t;
        x += lowbit(x);
    }
}

inline int Ask1(int x)
{
    int res = 0;
    while(x)
    {
        res += c_up[x];
        x -=lowbit(x);
    }
    return res;
}

inline int Ask2(int x)
{
    int res = 0;
    while(x)
    {
        res += c_down[x];
        x -= lowbit(x);
    }
    return res;
}

inline int query1(int l,int r)
{
    return Ask1(r) - Ask1(l-1);
}

inline int query2(int l,int r)
{
    return Ask2(r) - Ask2(l-1);
}

inline void dfs(int x,int sum)
{
    if(sum >= ans) return;
    if(x > num) 
    {
        ans = min(ans ,sum);
        return;    
    }
    add1(r[x],1);
    dfs(x+1,sum+min(query1(l[x],r[x]-1),query2(l[x],n)+query1(r[x]+1,n)));
    add1(r[x],-1);
    add2(r[x],1);
    dfs(x+1,sum+min(query2(l[x],r[x]-1),query2(r[x]+1,n)+query1(l[x],n)));    
    add2(r[x],-1);
}

int main()
{
    T = read();
    while(T--)
    {
        ans = 1<<30;num=0;
        memset(c_up,0,sizeof(c_up));
        memset(c_down,0,sizeof(c_down));
        n = read();
        for(int i=1;i<=n;i++) a[i] = read();
        for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) if(a[i] == a[j]) {l[++num] = i,r[num] = j;break;}
        dfs(1,0);
        printf("%d\n",ans);
    }
}