hdu 3473 划分树

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2721 Accepted Submission(s): 627



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

/*开成long long类型，超内存，部分用long long,部分用int就过了，看来long long 挺占内存空间的。
这题用划分树挺好做的。
题意：
一段固定不变的数字 ,m次询问,每次询问选择一个x值,使得区间[l,r]中每个元素与x的差的绝对值的和最小
思路：
x值明显选择[l,r]中数字的中位数,那么题目就变成了[l,r]中第(r-l+1+1)/2小的数是几,由于数字是静态的
,所以划分树可解
那么ans = num(<=x) * x - sum(<=x) + sum(>x) - num(>x) * x
由于sum之间可由前缀和相互求出  num也可以通过区间大小相互推出  因此只需要求 x 、 sum(<=x) 、 num(<=x)
最基本的划分树可以解决x和num的求解
想求sum可以维护一个lsum[i][j]数组  表示第i层到j位置为止进入左子树的数字的和*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 100005
typedef long long ll;
using namespace std;
int val[18]
,num[18]
,st
;
ll lsum[18]
,sum
,ans;
void build(int x,int y,int cen)
{
if(x==y)  return;
int mid=(x+y)>>1,same=mid-x+1,l=x,r=mid+1,i;
for(i=x; i<=y; i++)
if(val[cen][i]<st[mid])
same--;
for(i=x; i<=y; i++)
{
int flag=0;
if(val[cen][i]<st[mid]||(val[cen][i]==st[mid]&&same))
{
flag=1;
val[cen+1][l++]=val[cen][i];
lsum[cen][i]=lsum[cen][i-1]+val[cen][i];
if(val[cen][i]==st[mid])
same--;
}
else
{
val[cen+1][r++]=val[cen][i];
lsum[cen][i]=lsum[cen][i-1];
}
num[cen][i]=num[cen][i-1]+flag;
}
build(x,mid,cen+1);
build(mid+1,y,cen+1);
}
void init(int n)
{
int i;
sum[0]=0;
for(i=0; i<=17; i++)
{
val[i][0]=0;
num[i][0]=0;
lsum[i][0]=0;
}
for(i=1; i<=n; i++)
{
scanf("%d",&st[i]);
val[0][i]=st[i];
sum[i]=sum[i-1]+st[i];
}
sort(st+1,st+1+n);
build(1,n,0);
}
int query(int st,int ed,int k,int x,int y,int cen)
{
if(x==y)
return val[cen][x];
int mid=(x+y)>>1;
int lx=num[cen][st-1]-num[cen][x-1];
int ly=num[cen][ed]-num[cen][st-1];
int rx=st-1-x+1-lx;
int ry=ed-st+1-ly;
if(ly>=k)
return query(x+lx,x+lx+ly-1,k,x,mid,cen+1);
else
{
ans+=lsum[cen][ed]-lsum[cen][st-1];
st=mid+rx+1;
ed=mid+rx+ry;
return query(st,ed,k-ly,mid+1,y,cen+1);
}
}
int main()
{
int n,m,i,j,k,t,cnt=1,a,b,aa;
ll ans1,s;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
init(n);
scanf("%d",&m);
printf("Case #%d:\n",cnt++);
while(m--)
{
ans=0;  s=0;
scanf("%d%d",&i,&j);
i++;j++;
k=(j-i)/2+1;
aa=query(i,j,k,1,n,0);
a=k-1; b=j-i+1-k;
ans1=(sum[j]-sum[i-1])-ans-aa;
s+=(ll)aa*a-ans;
s+=ans1-(ll)aa*b;
printf("%I64d\n",s);
}
printf("\n");
}
return 0;
}