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


[align=left]Problem Description[/align]
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!

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2

5
3 6 2 2 4
2
1 4
0 2

2
7 7
2
0 1
1 1

[align=left]Sample Output[/align]

Case #1:
6
4

Case #2:
0
0

[align=left]Author[/align]
standy

[align=left]Source[/align]
2010 ACM-ICPC Multi-University Training Contest（4）——Host by UESTC

[align=left]Recommend[/align]
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:\n",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\n",ans);
}
printf("\n");
}
return 0;
}