hdu 3473 Minimum Sum（划分树）

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3566    Accepted Submission(s): 818

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

253 6 2 2 421 40 227 720 11 1

 

Sample Output

Case #1:64Case #2:00

 

n个数 m个查询 求区间[l,r]内 求和最小的值

满足求和最小的xi的值为该区间内的中位数

用sum[i][j]记录第i层的前缀和 

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int MAXN=200100;int tree[20][MAXN];int sorted[MAXN];int toleft[20][MAXN];long long sum[20][MAXN];void build(int l,int r,int dep){    if(l==r)    {        sum[dep][l]=tree[dep][l];        return;    }    int mid=(l+r)/2;    int same=mid-l+1;    for(int i=l;i<=r;i++)    {        if(tree[dep][i]<sorted[mid])            same--;        sum[dep][i]=tree[dep][i];        if(i>l)            sum[dep][i]+=sum[dep][i-1];    }    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];        else if(tree[dep][i]==sorted[mid]&&same>0)        {            tree[dep+1][lpos++]=tree[dep][i];            same--;        }        else            tree[dep+1][rpos++]=tree[dep][i];        toleft[dep][i]=toleft[dep][l-1]+lpos-l;    }    build(l,mid,dep+1);    build(mid+1,r,dep+1);}long long ans;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)/2;    int cnt=toleft[dep][r]-toleft[dep][l-1];    int ss=toleft[dep][l-1]-toleft[dep][L-1];    int ee=l-L-ss;    int s=toleft[dep][r]-toleft[dep][l-1];    int e=r-l+1-s;//l-r范围内没有进入左子树的个数    if(cnt>=k)    {        if(e>0)        {            if(ee>0) ans+=sum[dep+1][mid+e+ee]-sum[dep+1][mid+ee];            else ans+=sum[dep+1][mid+e];        }        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];        int newr=newl+cnt-1;        return query(L,mid,newl,newr,dep+1,k);    }    else    {        if(s>0)        {            if(ss>0) ans-=sum[dep+1][L+ss+s-1]-sum[dep+1][L+ss-1];            else ans-=sum[dep+1][L+s-1];        }        int newr=r+toleft[dep][R]-toleft[dep][r];        int newl=newr-(r-l-cnt);        return query(mid+1,R,newl,newr,dep+1,k-cnt);    }}int main(){    int tc;    scanf("%d",&tc);    int cs=1;    while(tc--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&tree[0][i]);            sorted[i]=tree[0][i];        }        sort(sorted+1,sorted+n+1);        build(1,n,0);        int m;        scanf("%d",&m);        printf("Case #%d:\n",cs++);        while(m--)        {            int s,t;            scanf("%d%d",&s,&t);            s++; t++;            ans=0;            int tt=query(1,n,s,t,0,(t-s)/2+1);            if((t-s+1)%2==0)            {                ans-=tt;            }            printf("%lld\n",ans);        }        puts("");    }    return 0;}