NOIP 模拟 $31\; \rm Time$

题解 \(by\;zj\varphi\)

考虑如何才能最优。

每次一定把当前最小值移动到边界上，那么看它向左还是向右移更优。

用树状数组维护一下即可，复杂度 \(\mathcal O\rm (nlogn)\)

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            ri f=1;x=0;register char ch=gc();
            while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();}
            while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
            return x=f?x:-x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    typedef long long ll;
    static const int N=1e5+7;
    int *a[N],vis[N],num[N],l[N],r[N],mx,n;
    ll ans;
    struct BIT{
        #define lowbit(x) ((x)&-(x)) 
        int c[N];
        inline void update(int x,int k) {for (ri i(x);i<=n;i+=lowbit(i)) c[i]+=k;}
        inline int query(int x) {
            int res(0);
            for (ri i(x);i;i-=lowbit(i)) res+=c[i];
            return res;
        }
    }B;
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        cin >> n;
        for (ri i(1);i<=n;p(i)) cin >> num[i],p(vis[num[i]]),mx=cmax(mx,num[i]);
        for (ri i(1);i<=mx;p(i)) {
            if (!vis[i]) continue;
            a[i]=new int[vis[i]+1];
            l[i]=1;
        }
        for (ri i(1);i<=n;p(i)) a[num[i]][p(r[num[i]])]=i,B.update(i,1);
        for (ri i(1);i<=mx;p(i)) {
            if (!vis[i]) continue;
            while(l[i]<=r[i]) {
                int x1=a[i][l[i]],x2=a[i][r[i]];
                int d1=B.query(x1-1),d2=B.query(n)-B.query(x2);
                if (d1<d2) {
                    ans+=d1;
                    B.update(x1,-1);
                    p(l[i]);
                } else {
                    ans+=d2;
                    B.update(x2,-1);
                    --r[i];
                }
                //printf("ans=%lld\n",ans);
            }
        }
        printf("%lld\n",ans);
        return 0;
    }
}
int main() {return nanfeng::main();}