Paper Plane Fly Away（2019hbcpc）

Topic links: https://ac.nowcoder.com/acm/contest/903/E

 

Problem-solving ideas: the girls like sequence mapped into a sequence girls sitting position, it can be seen just before the first required number i have a few large number, there are several smaller number behind is the answer, both seeking to reverse the problem . The sequence order of the array are sequentially inserted into the tree, with a weight of 1, utilizing the nature of the tree array, it can be quickly determined before the item i and, inserting positive and negative twice, the number can be determined to reverse all that is required.

 

AC Code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<cmath>
using namespace std;
#define ll long long
#define ull unsigned long long
#define ldb long double
#define lowbit(x) (x&-x)
#define db double
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define endl "\n"
#define rush() int T;scanf("%d",&T);while(T--)
#define mem(a,b) memset((a),(b),sizeof(a))
const db pi = acos((db)-1);
const ll MAXN = 1e5;
const ll mod = 1e9+7;

int a[MAXN+2],b[MAXN+2];
int s[MAXN+2];
int f[MAXN+2];

void add(int i, int x)
{
    while(i<MAXN)f[i]+=x,i+=lowbit(i);
}
int sum(int i)
{
    int res=0;
    while(i)res+=f[i],i-=lowbit(i);
    return res;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdin);
    
    mem(s,0);
    int n;
    scanf("%d",&n);
    
    for(int i=1;i<=n;i++)
    {
        int c;
        scanf("%d%d",&c,&b[i]);
        c-=n;
        a[c]=i;
        b[i]-=n;
    }
    
    for(int i=1;i<=n;i++)b[i]=a[b[i]];
    
    mem(f,0);
    for(int i=1;i<=n;i++)
    {
        add(b[i],1);
        s[i]+=sum(n+1)-sum(b[i]);
    }
    mem(f,0);
    for(int i=n;i>0;i--)
    {
        s[i]+=sum(b[i]);
        add(b[i],1);
    }
    
    for(int i=1;i<=n;i++)printf("%d\n",s[i]);
    
    return 0;
}