SPOJ GSS5 Can you answer these queries V

Description

You are given a sequence A[1], A[2], ..., A
. ( |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

又是查询最大连续字段和，但是限制了左右端点所在的区间……

线段树的部分不需要改动，计算答案的时候改一下即可。

如果区间有重复部分，就把区间分成三段，左段里找左端点，右段里找右端点，然后并上中段。有一串麻烦的判断，具体看代码。

如果区间没有重复部分，就左段里找左端点，右段里找右端点，然后强制加上两区间中间的序列和。

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define lc rt<<1
#define rc rt<<1|1
using namespace std;
const int mxn=100010;
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-'0';ch=getchar();}
return x*f;
}
int n,m;
int data[mxn];
struct node{
int mx;
int ml,mr;
int smm;
}t[mxn<<2],tmp0;
void Build(int l,int r,int rt){
if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;}
int mid=(l+r)>>1;
Build(l,mid,lc);
Build(mid+1,r,rc);
t[rt].smm=t[lc].smm+t[rc].smm;
t[rt].mx=max(t[lc].mx,t[rc].mx);
t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx);
t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml);
t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr);
return;
}
node query(int L,int R,int l,int r,int rt){
//    printf("%d %d %d %d %d\n",L,R,l,r,rt);
if(R<L){
return (node){0,0,0,0};
}
if(L<=l && r<=R){return t[rt];}
int mid=(l+r)>>1;
node res1;
if(L<=mid)res1=query(L,R,l,mid,lc);
else res1=tmp0;
node res2;
if(R>mid)res2=query(L,R,mid+1,r,rc);
else res2=tmp0;
node res={0};
res.smm=res1.smm+res2.smm;
res.mx=max(res1.mx,res2.mx);
res.mx=max(res.mx,res1.mr+res2.ml);
res.ml=max(res1.ml,res1.smm+res2.ml);
res.mr=max(res2.mr,res2.smm+res1.mr);
return res;
}
int qsum(int L,int R,int l,int r,int rt){
if(L<=l && r<=R)return t[rt].smm;
int mid=(l+r)>>1;
int res=0;
if(L<=mid)res+=qsum(L,R,l,mid,lc);
if(R>mid)res+=qsum(L,R,mid+1,r,rc);
return res;
}
int main(){
int T;
T=read();
while(T--){
n=read();
int i,j,x0,y0,x2,y2;
for(i=1;i<=n;i++)data[i]=read();
Build(1,n,1);
m=read();
tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=0;
for(i=1;i<=m;i++){
x0=read();y0=read();x2=read();y2=read();
int tmp=0;
int ans=-1e9;
if(y0>=x2){
//区间重叠
node res=query(x2,y0,1,n,1);
node res1=query(x0,x2-1,1,n,1);
node res2=query(y0+1,y2,1,n,1);
ans=max(ans,res1.mr+res.smm+res2.ml);
ans=max(ans,res1.mr+res.ml);
ans=max(ans,res.mr+res2.ml);
ans=max(ans,res.mx);
}
else{
//区间未重叠
if(y0+1<x2)tmp=qsum(y0+1,x2-1,1,n,1);
node res1=query(x0,y0,1,n,1);
node res2=query(x2,y2,1,n,1);
ans=max(ans,tmp+res1.mr+res2.ml);
}
//printf("%d\n",query(x2,y2,1,n,1).mx);
printf("%d\n",ans);
}
}
return 0;
}