hdu3473(划分树)

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2311    Accepted Submission(s): 540

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

 

       本题给定一个数的序列，然后给定若干查询区间，要找一个数使区间所有数与其绝对值的差最小，求这个最小绝对值的差。

          若单单是找这样的数，最好的当然是平均值，但若是找平均值本题就不好做了，于是我们可以考虑用中位数代替平均值，可以用几何线段解释。而找中位数是划分树的本职。其实可以在找中位数的过程中求得这样的最小值。

        我们可以开辟一个累加和数组sum[][],sum[dep][i]表示第dep层从1到i的累加和，ans为答案。当进入左子树查找时ans+=查询区间进入右子树的元素和；当进入右子树查询时ans-=查询区间进入左子树的元素和。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

const int MAXN=100000+100;
const int MAXPOW=20;//MAXPOW依据MAXN定大小
int tree[MAXPOW][MAXN];//tree[dep][i]表示第dep层第i个位置的值
__int64 sum[MAXPOW][MAXN];
int sorted[MAXN];//已经排序的数
int toleft[MAXPOW][MAXN];//toleft[dep][i]表示第dep层从1到i进入左边元素的个数
__int64 ans;

//构建深度为dep、区间为[l,r]的划分树
//时间复杂度为O(N*log(N))
void build(int l,int r,int dep)
{
	int i;
    int mid=(l+r)>>1;
    int same=mid-l+1;//表示等于中间元素而且被进入左边的元素个数
    for(i=l;i<=r;i++)
    {
		if(tree[dep][i]<sorted[mid])
			same--;
		sum[dep][i]+=sum[dep][i-1]+tree[dep][i];
	}
	if(l==r)return;
    int lpos=l;
    int rpos=mid+1;
    for(i=l;i<=r;i++)
    {
        if(tree[dep][i]<sorted[mid])//比中间的元素小，进入左边
			tree[dep+1][lpos++]=tree[dep][i];
        else if(tree[dep][i]==sorted[mid]&&same>0)
        {
            tree[dep+1][lpos++]=tree[dep][i];
            same--;
        }
        else  //比中间的元素大，进入右边
            tree[dep+1][rpos++]=tree[dep][i];
        toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的元素个数
    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
}


//查询小区间[l,r]内的第k大的元素，[L,R]是覆盖小区间的大区间
//时间复杂度为O(log(N))
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];//[l,r]中位于左边的元素个数
    if(cnt>=k)//进入左子树查询
    {
		int ee=r-L+1-(toleft[dep][r]-toleft[dep][L-1])+mid;
        int ss=l-L-(toleft[dep][l-1]-toleft[dep][L-1])+mid;
        ans+=sum[dep+1][ee]-sum[dep+1][ss];
        //修改小区间的l=L+要查询的区间前进入左边的元素个数
        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
        //r=newl+查询区间会被放在左边的元素个数
        int newr=newl+cnt-1;
        return query(L,mid,newl,newr,dep+1,k);
    }
    else//进入右子树查询
    {
		int s=L+toleft[dep][l-1]-toleft[dep][L-1];
        int e=s+cnt-1;
        ans-=sum[dep+1][e]-sum[dep+1][s-1];
		//修改小区间的r=r+要查询的区间后进入左边的元素个数
		int newr=r+toleft[dep][R]-toleft[dep][r];
		//l=r-要查询的区间进入右边的元素个数
		int newl=newr-(r-l-cnt);
		return query(mid+1,R,newl,newr,dep+1,k-cnt);
    }
}


int main()
{
    int n,m,i,l,r,k,tag=1;
	int cas;
	cin>>cas;
    while(cas--)
    {
		scanf("%d",&n);
		memset(sum,0,sizeof(sum));
        memset(tree,0,sizeof(tree));//这个必须
        for(i=1;i<=n;i++)//从1开始
        {
            scanf("%d",&tree[0][i]);
            sorted[i]=tree[0][i];
        }
        sort(sorted+1,sorted+n+1);
        build(1,n,0);
		scanf("%d",&m);
		printf("Case #%d:\n",tag++);
        while(m--)
        {
            scanf("%d%d",&l,&r);
			l++,r++;
			k=(r-l)/2+1;
			ans=0;
			
			int tmp = query(1,n,l,r,0,k);
            if((l+r)%2)ans-=tmp;
            printf("%I64d\n",ans);
			/*tmp=query(1,n,l,r,0,k);
			for(i=l;i<=r;i++)
			ans+=(__int64)abs(tree[0][i]-tmp);
			printf("%I64d\n",ans);*/
        }	
		printf("\n");
    }
    return 0;
}