hdu 3473 Minimum Sum

[align=left]Problem Description[/align]
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!

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2

5
3 6 2 2 4
2
1 4
0 2

2
7 7
2
0 1
1 1

[align=left]Sample Output[/align]

Case #1:
6
4

Case #2:
0
0

题意：给你一个区间[l,r],要求sum|x-xi|(l<=i<=r)的最小值,其中x必须为xl,xl+1...xr中的一个数。

分析：值最小时x为序列的中位数，可以用划分树求得，sum|x-xi|=sum(x-xi)(xi<=x)+sum(xi-x)(xi>x)，在划分树求中位数x的过程中可以计算出≤x的数（划分到左边）的个数sl和和suml,≥x的数（划分到右边）的个数sr和sumr，则ans=(sl-sr)*x+sumr-suml。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
const int N = 101010;
int sum[20]
,tr[20]
,sr
;
LL lsum, rsum;
LL lnum, rnum;
LL ss[20]
;
void build(int l, int r, int d)
{
for(int i=l; i<=r; i++) ss[d][i] = ss[d][i-1] + (LL)tr[d][i];
if(l == r) return;
int mid = l + r >> 1, i, lp = l, rp = mid+1;
int eqnum, tp=0;
for(int i=l; i<=r; i++) if(tr[d][i] < sr[mid]) tp++;
eqnum = mid-l+1-tp;
tp = 0;
for(i=l; i<=r; i++)
{
sum[d][i] = sum[d][i-1];
if(tr[d][i] < sr[mid])
tr[d+1][lp++] = tr[d][i], sum[d][i]++;
else if(tr[d][i] == sr[mid] && tp < eqnum)
tr[d+1][lp++] = tr[d][i], sum[d][i]++, tp++;
else tr[d+1][rp++] = tr[d][i];
}
build(l,mid,d+1);
build(mid+1,r,d+1);
}
int query(int s, int t, int k, int l, int r, int d)
{

if(s == t) return tr[d][s];
int mid = l + r >> 1;
int x = sum[d][s-1] - sum[d][l-1];
int y = sum[d][t] - sum[d][s-1];
if(y >= k) {
rsum += ss[d+1][mid+s-l-x+t-s+1-y]-ss[d+1][s-l-x+mid];
rnum += (mid + s-l-x + t-s+1-y)-(s-l-x+mid+1) + 1;
return query(l+x, l+x+y-1,k, l, mid, d+1);
}
lsum+=ss[d+1][l+x+y-1]-ss[d+1][l+x-1];
lnum+= (l+x+y-1) - (l+x) + 1;
return query(s-l-x+mid+1, mid + s-l-x + t-s+1-y, k-y, mid+1, r, d+1);
}
int main()
{
//freopen("date.in","r",stdin);
//freopen("date.out2","w",stdout);
int i,j,k,m,n;
int a,b,c;
int T;
int cas = 0;
for(i=0; i<20; i++) sum[i][0] = tr[i][0] = ss[i][0] = 0;
bool blank = false;
scanf("%d",&T);
while(T--)
{
printf("Case #%d:\n",++cas);
scanf("%d",&n);
for(i=1; i<=n; i++) scanf("%d",sr+i);
for(i=1; i<=n; i++) tr[0][i] = sr[i];
sort(sr+1, sr+1+n);
build(1,n,0);
scanf("%d",&m);
while(m--)
{
scanf("%d %d",&a,&b);
a++,b++;
if(a > b) swap(a,b);
int mid = a+b>>1;
k = mid-a+1;
lsum = rsum = 0;
lnum = rnum = 0;
c = query(a,b,k,1,n,0);
printf("%I64d\n",(lnum-rnum)*c+rsum-lsum);
}
puts("");
}
return 0;
}