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): 2235    Accepted Submission(s): 512

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

 

其实就是找到中间那个数。

然后需要记录和，右边的减掉左边的

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int MAXN = ;
int tree[][MAXN];
int sorted[MAXN];
int toleft[][MAXN];
long long sum[][MAXN];

void build(int l,int r,int dep)
{
    if(l == r)
    {
        sum[dep][l] = sum[dep][l-]+tree[dep][l];
        return;
    }
    int mid = (l+r)>>;
    int same = mid - l + ;
    for(int i = l;i <= r;i++)
    {
        if(tree[dep][i] < sorted[mid])
            same--;
        sum[dep][i] += sum[dep][i-]+tree[dep][i];
    }
    int lpos = l;
    int rpos = mid+;
    for(int i = l;i <= r;i++)
    {
        if(tree[dep][i] < sorted[mid])
            tree[dep+][lpos++] = tree[dep][i];
        else if(tree[dep][i] == sorted[mid] && same > )
        {
            tree[dep+][lpos++] = tree[dep][i];
            same--;
        }
        else
            tree[dep+][rpos++] = tree[dep][i];
        toleft[dep][i] = toleft[dep][l-] + lpos - l;
    }
    build(l,mid,dep+);
    build(mid+,r,dep+);
}
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)>>;
    int cnt = toleft[dep][r] - toleft[dep][l-];
    if(cnt >= k)
    {
        int ee = r-L+-(toleft[dep][r]-toleft[dep][L-])+mid;
        int ss = l-L-(toleft[dep][l-]-toleft[dep][L-])+mid;

        ans += sum[dep+][ee]-sum[dep+][ss];

        int newl = L + toleft[dep][l-]-toleft[dep][L-];
        int newr = newl + cnt -;
        return query(L,mid,newl,newr,dep+,k);
    }
    else
    {
        int s = L + toleft[dep][l-] - toleft[dep][L-];
        int e = s + cnt - ;

        ans -= sum[dep+][e] - sum[dep+][s-];

        int newr = r + toleft[dep][R] - toleft[dep][r];
        int newl = newr - (r-l+-cnt) + ;
        return query(mid+,R,newl,newr,dep+,k-cnt);
    }
}
int main()
{
    int T;
    int n;
    scanf("%d",&T);
    int iCase = ;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        memset(tree,,sizeof(tree));
        memset(sum,,sizeof(sum));
        for(int i = ;i <= n;i++)
        {
            scanf("%d",&tree[][i]);
            sorted[i] = tree[][i];
        }
        sort(sorted+,sorted+n+);
        build(,n,);
        printf("Case #%d:\n",iCase);
        int m,l,r;
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d",&l,&r);
            l++;r++;
            ans = ;
            int tmp = query(,n,l,r,,(l+r)/-l+);
            if((l+r)%)ans-=tmp;
            printf("%I64d\n",ans);
        }
        printf("\n");
    }
    return ;
}