hdu 3473 Minimum Sum(划分树应用)

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1620    Accepted Submission(s): 379

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

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3473

题意：给你一个数列，每次询问一个区间，要求找到一个数，使得区间所有数与它的差的绝对值之和最小。。。

分析：这里涉及到一个常识，那个数就是区间的中位数，那么原问题就转成，询问一个区间的中位数与所有数的差，可以用划分树解决。。。

具体的实现方案，只要添加一个当前小于中间值的数的和即可，询问时再做统计，最后根据简单的加减就能算出答案了

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int mm=111111;
int a[mm],s[22][mm],g[22][mm];
long long w[22][mm],sum[mm],tmp,ans;
int i,j,k,n,m,t,cs=0;
void build(int l,int r,int d)
{
    if(l==r)return;
    int i,t1,t2,same=0,m=(l+r)>>1;
    for(i=l; i<=r; ++i)
        w[d][i]=g[d][i]=0,same+=(s[d][i]<a[m]);
    same=m-l+1-same;
    for(t1=i=l,t2=m+1; i<=r; ++i)
        if(s[d][i]==a[m]&&same)--same,s[d+1][t1++]=s[d][i],g[d][i]=1,w[d][i]=s[d][i];
        else if(s[d][i]<a[m])
            s[d+1][t1++]=s[d][i],g[d][i]=1,w[d][i]=s[d][i];
        else s[d+1][t2++]=s[d][i];
    for(i=l; i<r; ++i)
        g[d][i+1]+=g[d][i],w[d][i+1]+=w[d][i];
    build(l,m,d+1);
    build(m+1,r,d+1);
}
int query(int L,int R,int k,int l,int r,int d)
{
    int m,w1,w2;
    long long tmp;
    while(l<r)
    {
        m=(l+r)>>1,w1=0,w2=g[d][R],tmp=0;
        if(L>l)w2-=(w1=g[d][L-1]),tmp=w[d][L-1];
        if(w2>=k)r=m;
        else ans-=(w[d][R]-tmp),
        k-=w2,w1=L-l-w1,w2=R-L+1-w2,l=m+1;
        L=l+w1,R=l+w1+w2-1,++d;
    }
    return s[d][l];
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1; i<=n; ++i)
        {
            scanf("%d",&a[i]);
            s[0][i]=a[i];
            sum[i]=sum[i-1]+a[i];
        }
        sort(a+1,a+n+1);
        build(1,n,0);
        scanf("%d",&m);
        printf("Case #%d:\n",++cs);
        while(m--)
        {
            scanf("%d%d",&i,&j);
            ++i,++j;
            k=(j-i+2)/2;
            ans=0;
            tmp=query(i,j,k,1,n,0);
            ans=sum[j]-sum[i-1]+2*ans-tmp*(j-i+2-k)+tmp*(k-1);
            printf("%I64d\n",ans);
        }
        puts("");
    }
    return 0;
}