SPOJ GSS5 Can you answer these queries V

Time Limit: 132MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

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.

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=;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,m;
 int data[mxn];
 struct node{
     int mx;
     int ml,mr;
     int smm;
 }t[mxn<<],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)>>;
     Build(l,mid,lc);
     Build(mid+,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){,,,};
     }
     if(L<=l && r<=R){return t[rt];}
     int mid=(l+r)>>;
     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+,r,rc);
         else res2=tmp0;
     node res={};
     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)>>;
     int res=;
     if(L<=mid)res+=qsum(L,R,l,mid,lc);
     if(R>mid)res+=qsum(L,R,mid+,r,rc);
     return res;
 }
 int main(){
     int T;
     T=read();
     while(T--){
         n=read();
         int i,j,x0,y0,x2,y2;
         for(i=;i<=n;i++)data[i]=read();
         Build(,n,);
         m=read();
         tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=;
         for(i=;i<=m;i++){
             x0=read();y0=read();x2=read();y2=read();
             int tmp=;
             int ans=-1e9;
             if(y0>=x2){
             //区间重叠
                 node res=query(x2,y0,,n,);
                 node res1=query(x0,x2-,,n,);
                 node res2=query(y0+,y2,,n,);
                 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+<x2)tmp=qsum(y0+,x2-,,n,);
                 node res1=query(x0,y0,,n,);
                 node res2=query(x2,y2,,n,);
                 ans=max(ans,tmp+res1.mr+res2.ml);
             }
             //printf("%d\n",query(x2,y2,1,n,1).mx);
             printf("%d\n",ans);
         }
     }
     return ;
 }