BZOJ4240 有趣的家庭菜园(贪心+树状数组)

  显然相当于使序列变成单峰。给原序列每个数按位置标号,则要求重排后的序列原标号的逆序对数最少。考虑将数从大到小放进新序列,那么贪心的考虑放在左边还是右边即可,因为更小的数一定会在其两侧,与它自身放在哪无关。对于相同的数,一定可以将其安排至之间无逆序对,特判一下。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 300010
int n,tree[N];
long long ans=0;
struct data{int x,i;
}a[N];
bool cmp(const data&a,const data&b)
{
    return a.x>b.x;
}
void add(int k){while (k<=n) tree[k]++,k+=k&-k;}
int query(int k){int s=0;while (k) s+=tree[k],k-=k&-k;return s;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4240.in","r",stdin);
    freopen("bzoj4240.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i].x=read(),a[i].i=i;
    sort(a+1,a+n+1,cmp);
    for (int i=1;i<=n;i++)
    {
        int t=i;while (t<n&&a[t+1].x==a[i].x) t++;
        for (int j=i;j<=t;j++) ans+=min(query(a[j].i),i-1-query(a[j].i));
        for (int j=i;j<=t;j++) add(a[j].i);
        i=t;
    }
    cout<<ans;
    return 0;
}

 

posted @ 2018-10-22 23:26  Gloid  阅读(198)  评论(0编辑  收藏  举报