SPOJ GSS1

Can you answer these queries I SPOJ - GSS1
You are given a sequence A[1], A[2], …, A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows:
Query(x,y) = Max { a[i]+a[i+1]+…+a[j] ; x ≤ i ≤ j ≤ y }.
Given M queries, your program must output the results of these queries.
Input
The first line of the input file contains the integer N.
In the second line, N numbers follow.
The third line contains the integer M.
M lines follow, where line i contains 2 numbers xi and yi.
Output
Your program should output the results of the M queries, one query per line.
Example
Input:
3
-1 2 3
1
1 2
Output:
2

/*
一开始W.
不知道为啥.
拍了好多组数据都OK.
原来case更新的时候错了.
考虑三种情况.
分别维护GSS,LGSS,RGSS.
分为两种形态:跨区间和不跨区间. 
case 1,2:左右段的GSS.
case 3:左段右端与右段左端的GSS和.
一开始更新的时候更新成了该段的左端GSS 右端GSS case3.
画了画图不对吖.
如果跨区间的话这两种情况是包含在case3里边的.
然后这样就忽略了case1,2. 
*/
#include<iostream>
#include<cstdio>
#define MAXN 50001
using namespace std;
int n,m,cut;
struct data{
    int l,r,lg,rg,g,sum,size;
    data *lc,*rc;
}tree[MAXN*4];
int read()
{
    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-48,ch=getchar();
    return x*f;
}
void build(data *k,int l,int r,int now)
{
    k->l=l,k->r=r;
    if(l==r) {k->g=k->lg=k->rg=k->sum=read();return ;}
    int mid=(l+r)>>1;
    k->size=now;
    k->lc=&tree[now*2];
    k->rc=&tree[now*2+1];
    k->lc->size=now*2;
    k->rc->size=now*2+1;
    build(k->lc,l,mid,now*2);
    build(k->rc,mid+1,r,now*2+1);
    k->lg=max(k->lc->lg,k->lc->sum+k->rc->lg);
    k->rg=max(k->rc->rg,k->rc->sum+k->lc->rg);
    k->sum=k->lc->sum+k->rc->sum;
    k->g=max(k->lc->g,max(k->lc->rg+k->rc->lg,k->rc->g));
    return ;
}
data query(data *k,int l,int r,int num)
{
    data xx;
    if(l<=k->l&&k->r<=r) return tree[num];
    int mid=(k->l+k->r)>>1;
    if(l>mid) return query(k->rc,l,r,k->rc->size);
    else if(r<=mid) return query(k->lc,l,r,k->lc->size);
    else {
        data ll=query(k->lc,l,mid,k->lc->size);
        data rr=query(k->rc,mid+1,r,k->rc->size);
        xx.sum=ll.sum+rr.sum;
        xx.lg=max(ll.lg,ll.sum+rr.lg);
        xx.rg=max(rr.rg,rr.sum+ll.rg);
        xx.g=max(ll.g,max(rr.g,ll.rg+rr.lg));
    }
    return xx;
}
int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int x,y;
    n=read();
    build(tree+1,1,n,1);
    m=read();
    while(m--)
    {
        x=read(),y=read();
        data xx=query(tree+1,x,y,1);
        printf("%d
",xx.g);
    }
    return 0;
}