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 = 100010;
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] = sum[dep][l-1]+tree[dep][l];
        return;
    }
    int mid = (l+r)>>1;
    int same = mid - l + 1;
    for(int i = l;i <= r;i++)
    {
        if(tree[dep][i] < sorted[mid])
            same--;
        sum[dep][i] += sum[dep][i-1]+tree[dep][i];
    }
    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)>>1;
    int cnt = toleft[dep][r] - toleft[dep][l-1];
    if(cnt >= k)
    {
        int ee = r-L+1-(toleft[dep][r]-toleft[dep][L-1])+mid;
        int ss = l-L-(toleft[dep][l-1]-toleft[dep][L-1])+mid;

        ans += sum[dep+1][ee]-sum[dep+1][ss];

        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
    {
        int s = L + toleft[dep][l-1] - toleft[dep][L-1];
        int e = s + cnt - 1;

        ans -= sum[dep+1][e] - sum[dep+1][s-1];

        int newr = r + toleft[dep][R] - toleft[dep][r];
        int newl = newr - (r-l+1-cnt) + 1;
        return query(mid+1,R,newl,newr,dep+1,k-cnt);
    }
}
int main()
{
    int T;
    int n;
    scanf("%d",&T);
    int iCase = 0;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        memset(tree,0,sizeof(tree));
        memset(sum,0,sizeof(sum));
        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);
        printf("Case #%d:
",iCase);
        int m,l,r;
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d",&l,&r);
            l++;r++;
            ans = 0;
            int tmp = query(1,n,l,r,0,(l+r)/2-l+1);
            if((l+r)%2)ans-=tmp;
            printf("%I64d
",ans);
        }
        printf("
");
    }
    return 0;
}