SPOJ GSS5 Can you answer these queries V ——线段树

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>

#include <map>
#include <set>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;

#define maxn 500005
#define eps 1e-8
#define db double
#define ll long long
#define inf 0x3f3f3f3f
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)

void Finout()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    #endif
}

int Getint()
{
    int x=0,f=1; char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) f=-1; ch=getchar();}
    while (ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}

struct Node{
	int lx,rx,mx,sum;
	Node operator + (Node x)
	{
		Node ret;
		ret.lx=max(lx,sum+x.lx);
		ret.rx=max(x.sum+rx,x.rx);
		ret.sum=sum+x.sum;
		ret.mx=max(max(mx,x.mx),max(rx+x.lx,max(ret.lx,ret.rx)));
		return ret;
	}
}t[maxn];

int n,a[maxn],q,L,R;

void Build(int o,int l,int r)
{
//	printf("Build %d %d\n",l,r);
	if (l==r)
	{
		t[o].lx=t[o].rx=t[o].mx=t[o].sum=a[l];
		return ;
	}
	int mid=l+r>>1;
	Build(o<<1,l,mid);
	Build(o<<1|1,mid+1,r);
	t[o]=t[o<<1]+t[o<<1|1];
}

Node Query(int o,int l,int r)
{
//	printf("Query %d %d\n",l,r); 
	if (L<=l&&r<=R) return t[o];
	int mid=l+r>>1;
	if (L>mid) return Query(o<<1|1,mid+1,r);
	else if (R<=mid) return Query(o<<1,l,mid);
	else return Query(o<<1,l,mid)+Query(o<<1|1,mid+1,r);
}

int main()
{
	int T;
	Finout();scanf("%d",&T);
	while (T--)
	{
		memset(a,0,sizeof a);
		memset(t,0,sizeof t);
		scanf("%d",&n);
		F(i,1,n) scanf("%d",&a[i]);
		Build(1,1,n);
		scanf("%d",&q);
		F(i,1,q)
		{
			int x1=Getint(),y1=Getint(),x2=Getint(),y2=Getint(),ans=0;
			if (x2>y1)
			{
				L=x1;R=y1; ans+=Query(1,1,n).rx;// printf("ans 1: %d\n",ans);
				L=x2;R=y2; ans+=Query(1,1,n).lx; //printf("ans 2: %d\n",ans);
				L=y1+1;R=x2-1; if (R>=L) ans+=Query(1,1,n).sum;//printf("ans 3: %d\n",ans);
			}
			else
			{
				int tmp=0;
				L=x2;R=y1; ans=Query(1,1,n).mx;
				if (x2>x1)
				{L=x1;R=x2-1;tmp+=Query(1,1,n).rx;}
				L=x2;R=y2; tmp+=Query(1,1,n).lx;
				ans=max(ans,tmp); tmp=0;
				L=x2;R=y1; tmp+=Query(1,1,n).rx;
				if (y2>y1)
				{L=y1+1;R=y2;tmp+=Query(1,1,n).lx;}
				ans=max(ans,tmp);
			}
			printf("%d\n",ans);
		}
	}
}