HDU 3473 Minimum Sum（划分树，求中位数，小于中位数的和与大于中位数的和）

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1533    Accepted Submission(s): 345

Problem Description

You are given N positive integers, denoted as x0, x1 ... xN-1. Then give you some intervals [l, r]. For each interval, you need to find a number x to make as small as possible!

 

 

Input

The first line is an integer T (T <= 10), indicating the number of test cases. For each test case, an integer N (1 <= N <= 100,000) comes first. Then comes N positive integers x (1 <= x <= 1,000, 000,000) in the next line. Finally, comes an integer Q (1 <= Q <= 100,000), indicting there are Q queries. Each query consists of two integers l, r (0 <= l <= r < N), meaning the interval you should deal with.

 

 

Output

For the k-th test case, first output “Case #k:” in a separate line. Then output Q lines, each line is the minimum value of . Output a blank line after every test case.

 

 

Sample Input

2 5 3 6 2 2 4 2 1 4 0 2 2 7 7 2 0 1 1 1

 

 

Sample Output

Case #1: 6 4 Case #2: 0 0

 

 

Author

standy

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（4）——Host by UESTC

 

 

Recommend

zhengfeng

 

 

 

 

这题就是划分树。

很容易知道当x为中位数时，题目中的式子最小。

增加个sum[i][j]来记录第i层1-j中数的和。

然后求比中位数大的数的和-比中位数小的数的和时，要注意，具体看代码。

/*
HDU 3473  Minimum Sum
AC  G+++  546ms  29000K

*/


#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=200010;
int tree[20][MAXN];
int sorted[MAXN];
int toleft[20][MAXN];
long long sum[20][MAXN];

void build(int l,int r,int dep)
{
    if(l==r)
    {
        sum[dep][l]=tree[dep][l];
        return;
    }
    int mid=(l+r)>>1;
    int same=mid-l+1;//same表示等于中间值且到左边的数的个数
    for(int i=l;i<=r;i++)
    {
        if(tree[dep][i]<sorted[mid])
        same--;
        sum[dep][i]=tree[dep][i];
        if(i>l)sum[dep][i]+=sum[dep][i-1];
    }

    int lpos=l;
    int rpos=mid+1;
    for(int i=l;i<=r;i++)
    {

        if(tree[dep][i]<sorted[mid])//去左边
        {
            tree[dep+1][lpos++]=tree[dep][i];
        }
        else if(tree[dep][i]==sorted[mid]&&same>0)//去左边
        {
            tree[dep+1][lpos++]=tree[dep][i];
            same--;

        }
        else//去右边
           tree[dep+1][rpos++]=tree[dep][i];
       toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的个数
    }
    build(l,mid,dep+1);//递归建树
    build(mid+1,r,dep+1);
}



long long ans;
int query(int L,int R,int l,int r,int dep,int k)
{
    if(l==r)return tree[dep][l];
    int mid=(L+R)>>1;
    int cnt=toleft[dep][r]-toleft[dep][l-1];

    int ss=toleft[dep][l-1]-toleft[dep][L-1];
    int ee=l-L-ss;
    int s=toleft[dep][r]-toleft[dep][l-1];
    int e=r-l+1-s;



    if(cnt>=k)
    {

        if(e>0)
        {
            if(ee>0)ans+=sum[dep+1][mid+e+ee]-sum[dep+1][mid+ee];
            else ans+=sum[dep+1][mid+e];
        }

        //L+查询区间前去左边的数的个数
        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
        //左端点+查询区间会分入左边的数的个数
        int newr=newl+cnt-1;
        return query(L,mid,newl,newr,dep+1,k);//注意
    }
    else
    {

        if(s>0)
        {
            if(ss>0)ans-=sum[dep+1][L+ss+s-1]-sum[dep+1][L+ss-1];
            else ans-=sum[dep+1][L+s-1];
        }

        //r+区间后分入左边的数的个数
        int newr=r+toleft[dep][R]-toleft[dep][r];
        //右端点减去区间分入右边的数的个数
        int newl=newr-(r-l-cnt);
        return query(mid+1,R,newl,newr,dep+1,k-cnt);//注意
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    int T;
    int n,m;
    scanf("%d",&T);
    int l,r;
    int iCase=0;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&tree[0][i]);
            sorted[i]=tree[0][i];
        }
        sort(sorted+1,sorted+1+n);
        build(1,n,0);



        printf("Case #%d:\n",iCase);
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d",&l,&r);
            l++;
            r++;
            ans=0;
            int tt=query(1,n,l,r,0,(r-l)/2+1);
            if((r-l+1)%2==0)
            {
                ans-=tt;
            }
            printf("%I64d\n",ans);
        }
        printf("\n");
    }
    return 0;
}