[树状数组] BZOJ 4240 有趣的家庭菜园

%%%POPOQQQ

从小到大枚举高度，由于无论将这株草移动到左侧还是右侧都对比它高的植物没有影响，因此贪心选择代价最小的方向即可

故答案为∑min(a[1…i-1]中大于a[i]的数的数量,a[i+1…n]中大于a[i]的数的数量)

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;

inline char nc()
{
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x)
{
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=300005;

struct BIT{
int maxn;
int c
;
#define lowbit(x) ((x)&-(x))
inline void init(int n){
maxn=n;
}
inline void add(int x,int r){
for (int i=x;i<=maxn;i+=lowbit(i))
c[i]+=r;
}
inline int sum(int x){
int ret=0;
for (int i=x;i;i-=lowbit(i))
ret+=c[i];
return ret;
}
inline int sum(int l,int r){
return sum(r)-sum(l-1);
}
}bit1,bit2;

int n,a
;
int sx
,icnt;

inline int Bin(int x){
return lower_bound(sx+1,sx+icnt+1,x)-sx;
}

int main()
{
ll ans=0;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n);
for (int i=1;i<=n;i++)
read(a[i]),sx[++icnt]=a[i];
sort(sx+1,sx+icnt+1);
icnt=unique(sx+1,sx+icnt+1)-sx-1;
bit1.init(icnt); bit2.init(icnt);
for (int i=1;i<=n;i++)
{
a[i]=Bin(a[i]);
bit2.add(a[i],1);
}
for (int i=1;i<=n;i++)
{
bit2.add(a[i],-1);
ans+=min(bit1.sum(a[i]+1,icnt),bit2.sum(a[i]+1,icnt));
bit1.add(a[i],1);
}
printf("%lld\n",ans);
return 0;
}