[树状数组] Jzoj P4737 金色丝线将瞬间一分为二

Description

 

Input

Output

 

Sample Input

5 10
1 1
2 2
3 3
4 4
5 5

Sample Output

4

 

Data Constraint

 

Hint

 

 

题解

• 先定义两个数组rank1和rank2表示以x或y坐标从小到大排序后第i个点的排名
• 那么弄出这两个数组之后，每次加上做到的点i到所有前面的点的距离
• 可以用两个树状数组来做，一个记录的是x坐标一个是y坐标
• 那么，就是先将1~rank1[i]排名的遗体的个数和x坐标和求出来，总距离加上num*x[i]-sumx
• 然后还有求出rank1[i]+1~n排名的遗体的个数和x坐标和求出来，总距离就是sumx-num*x[i]
• y的做法同上

代码

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
long long sz1[600010],sz2[600010],x[600010],y[600010],s,num,sumx,sumy,d;
int n,cnt1[600010],cnt2[600010],rank1[600010],rank2[600010],d1[600010],d2[600010];
bool cmpx(int a,int b) { return x[a]<x[b]; }
bool cmpy(int a,int b) { return y[a]<y[b]; }
void insert1(int x,int y) 
{ 
    for (;x<=n;x+=x&-x) sz1[x]+=y,cnt1[x]++; 
}
void insert2(int x,int y) 
{ 
    for (;x<=n;x+=x&-x) sz2[x]+=y,cnt2[x]++; 
}
long long getsum1(int x,long long &y) 
{ 
    long long s=0;
    for (;x;x-=x&-x) s+=sz1[x],y+=cnt1[x];
    return s;
}
long long getsum2(int x,long long &y) 
{ 
    long long s=0;
    for (;x;x-=x&-x) s+=sz2[x],y+=cnt2[x];
    return s;
}
long long query1(int l,int r,long long &num)
{
    long long tot=0,s=getsum1(r,num)-getsum1(l-1,tot);
    num-=tot;
    return s;
}
long long query2(int l,int r,long long &num)
{
    long long tot=0,s=getsum2(r,num)-getsum2(l-1,tot);
    num-=tot;
    return s;
}
int main()
{
    scanf("%d%lld",&n,&d);
    for (int i=1;i<=n;i++)
    {
        scanf("%lld%lld",&x[i],&y[i]);
        d1[i]=d2[i]=i;
    }
    sort(d1+1,d1+n+1,cmpx);
    sort(d2+1,d2+n+1,cmpy);
    for (int i=1;i<=n;i++) rank1[d1[i]]=i,rank2[d2[i]]=i;
    for (int i=1;i<=n;i++)
    {
        num=0,sumx=query1(1,rank1[i],num),s+=num*x[i]-sumx;
        num=0,sumx=query1(rank1[i],n,num),s+=sumx-num*x[i];
        num=0,sumy=query2(1,rank2[i],num),s+=num*y[i]-sumy;
        num=0,sumy=query2(rank2[i],n,num),s+=sumy-num*y[i];
        if (s>d)
        {
            printf("%d",i);
            return 0;
        }
        insert1(rank1[i],x[i]),insert2(rank2[i],y[i]);
    }
    printf("-1");
    return 0;
}