hdu 3473 划分树

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3611    Accepted Submission(s): 829

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

题意：

给你一个区间，让你找出其中一个值，让所有值与它相减的绝对值的和最小。

思路：

于是成了找出其中的中间值减去最小，最开始简单粗暴地超时。然后发现求绝对值有点麻烦，于是在建树和查找的过程中找出比中间值小的数的和，然后分类计算吧。

感觉最开始没有考虑好方案，太随意了。

/*
hdu 3473
最开始直接算TL，然后发现由于是计算绝对值，在建树的时候需要记录比你查找的值小的和，
然后分开计算即可
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <Map>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;

const int maxn = 100010;

int tree[20][maxn];
int sorted[maxn];
int toleft[20][maxn];
ll sum[maxn];
ll lsum[20][maxn];
ll tsum;

void build(int l,int r,int dep)  //模拟快排 并记录左树中比i小的个数
{
    if(l == r)
        return;
    int mid = (l+r)>>1;
    int same = mid-l+1;//可能有很多值与中间那个相等，但不一定被分到左边
    for(int i = l;i <= r;i++)
    {
        if(tree[dep][i] < sorted[mid])
            same--;
    }
    int lpos = l;
    int rpos = mid+1;
    for(int i = l;i <= r;i++)
    {
        if(tree[dep][i] < sorted[mid]){
            tree[dep+1][lpos++] = tree[dep][i];
            lsum[dep][i] = lsum[dep][i-1] + tree[dep][i];
        }
        else if(tree[dep][i] == sorted[mid] && same > 0)
        {
            tree[dep+1][lpos++] = tree[dep][i];
            lsum[dep][i] = lsum[dep][i-1] + tree[dep][i];
            same --;
        }
        else
        {
            tree[dep+1][rpos++] = tree[dep][i];
            lsum[dep][i] = lsum[dep][i-1];
        }
        toleft[dep][i] = toleft[dep][l-1] + lpos -l;

    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
 }


 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];  //所查找区间放在左树中的个数
     if(cnt >= k)
     {   //得到l左边放到左子树的个数，加上L即是开始位置
         int lpos = L+toleft[dep][l-1]-toleft[dep][L-1];
         int rpos = lpos+cnt-1;

         return query(L,mid,lpos,rpos,dep+1,k);
     }
     else
     {   //R-r可以得出后面空出了多少位置
         int rpos = r+toleft[dep][R]-toleft[dep][r];
         int lpos = rpos-(r-l-cnt);
         tsum += lsum[dep][r] - lsum[dep][l-1];
         return query(mid+1,R,lpos,rpos,dep+1,k-cnt);
     }
 }


 int main()
 {
     int n,m,T;
     int cas = 1;
     scanf("%d",&T);
     while(T--)
     {

         scanf("%d",&n);
         for(int i = 1;i <= n;i++)
         {
             scanf("%d",&sorted[i]);
             sum[i] = sum[i-1]+sorted[i];
             tree[0][i] = sorted[i];
         }
         sort(sorted+1,sorted+n+1);
         build(1,n,0);
         scanf("%d",&m);
         int l,r;
         printf("Case #%d:
",cas++);
         while(m--)
         {
             scanf("%d%d",&l,&r);
             l++;r++;tsum = 0;
             ll k =(r-l)/2+1;
             ll ans = 0;
             ll aver = query(1,n,l,r,0,k);
             ans = sum[r]-sum[l-1]-aver*(r-l+1-k)-tsum-aver;
             ans += (k-1)*aver-tsum;
             printf("%I64d
",ans);
         }
         puts("");
     }
     return 0;
 }