SPOJ Can you answer these queries V

Description

You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| <= 10000 , 1 <= N <= 10000 ). A query is defined as follows: Query(x1,y1,x2,y2) = Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 <= j <= y2 and x1 <= x2 , y1 <= y2 }. Given M queries (1 <= M <= 10000), your program must output the results of these queries.

Input

The first line of the input consist of the number of tests cases <= 5. Each case consist of the integer N and the sequence A. Then the integer M. M lines follow, contains 4 numbers x1, y1, x2 y2.

Output

Your program should output the results of the M queries for each test case, one query per line.

Example

Input:
2
6 3 -2 1 -4 5 2
2
1 1 2 3
1 3 2 5
1 1
1
1 1 1 1

Output:
2
3
1

Hint

Added by:	Frank Rafael Arteaga
Date:	2008-08-06
Time limit:	0.132s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Languages:	All except: C99 strict ERL JS NODEJS PERL 6 VB.net
Resource:	K.-Y. Chen and K.-M. Chao, On the Range Maximum-Sum Segment Query Problem, 2007.

一道吼题。线段树求区间最大子段和的变形。

有两种情况，区间分离或有重合。

情况太多，并不想一一列举，看代码吧。

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
const int N=50005;
int n,m,T;
long long data[N];
struct node
{
	int l,r;
	long long sum,tot,lm,rm;
}a[4*N];
void build(int num,int l,int r)
{
	a[num].l=l,a[num].r=r;
	if(l==r)
	{
		a[num].sum=a[num].tot=a[num].lm=a[num].rm=data[l];
		return ;
	}
	int mid=(l+r)/2;
	build(2*num,l,mid);
	build(2*num+1,mid+1,r);
	a[num].sum=a[2*num].sum+a[2*num+1].sum;
	a[num].lm=max(a[2*num].lm,a[2*num].sum+a[2*num+1].lm);
	a[num].rm=max(a[2*num+1].rm,a[2*num+1].sum+a[2*num].rm);
	a[num].tot=max(max(a[2*num].tot,a[2*num+1].tot),a[2*num].rm+a[2*num+1].lm);
}
node query(int num,int l,int r)
{
	if(l>r)
	{
		node u;
		u.lm=u.rm=u.sum=u.tot=0;
		return u;
	}
	if(a[num].l>=l&&a[num].r<=r)
		return a[num];
	if(a[2*num].r>=r)
		return query(2*num,l,r);
	if(a[2*num+1].l<=l)
		return query(2*num+1,l,r);
	node u1,u2,res;
	u1=query(2*num,l,r);
	u2=query(2*num+1,l,r);
	res.sum=u1.sum+u2.sum;
	res.lm=max(u1.lm,u1.sum+u2.lm);
	res.rm=max(u2.rm,u2.sum+u1.rm);
	res.tot=max(max(u1.tot,u2.tot),u1.rm+u2.lm);
	return res;
}
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
			scanf("%lld",&data[i]);
		build(1,1,n);
		scanf("%d",&m);
		int x1,y1,x2,y2;
		while(m--)
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			if(y1<x2)
				printf("%lld\n",query(1,y1,x2).sum+max(0ll,query(1,x1,y1-1).rm)+max(0ll,query(1,x2+1,y2).lm));
			else
			{
				long long t=query(1,x2,y1).tot;
				t=max(t,query(1,x2,y1).lm+max(0ll,query(1,x1,x2-1).rm));
				t=max(t,query(1,x2,y1).rm+max(0ll,query(1,y1+1,y2).lm));
				t=max(t,query(1,x2,y1).sum+max(0ll,query(1,x1,x2-1).rm)+max(0ll,query(1,y1+1,y2).lm));
				printf("%lld\n",t);
			}
		}
	}
	return 0;
}