SPOJ GSS5 Can you answer these queries V

Posted 2020-08-21 SilverNebula

 

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。

  1 /*by SilverN*/
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<cstdio>
  6 #include<cmath>
  7 #define lc rt<<1
  8 #define rc rt<<1|1
  9 using namespace std;
 10 const int mxn=100010;
 11 int read(){
 12     int x=0,f=1;char ch=getchar();
 13     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 14     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
 15     return x*f;
 16 }
 17 int n,m;
 18 int data[mxn];
 19 struct node{
 20     int mx;
 21     int ml,mr;
 22     int smm;
 23 }t[mxn<<2],tmp0;
 24 void Build(int l,int r,int rt){
 25     if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;}
 26     int mid=(l+r)>>1;
 27     Build(l,mid,lc);
 28     Build(mid+1,r,rc);
 29     t[rt].smm=t[lc].smm+t[rc].smm;
 30     t[rt].mx=max(t[lc].mx,t[rc].mx);
 31     t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx);
 32     t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml);
 33     t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr);
 34     return;
 35 }
 36 node query(int L,int R,int l,int r,int rt){
 37 //    printf("%d %d %d %d %d\n",L,R,l,r,rt);
 38     if(R<L){
 39         return (node){0,0,0,0};
 40     }
 41     if(L<=l && r<=R){return t[rt];}
 42     int mid=(l+r)>>1;
 43     node res1;
 44     if(L<=mid)res1=query(L,R,l,mid,lc);
 45         else res1=tmp0;
 46     node res2;
 47     if(R>mid)res2=query(L,R,mid+1,r,rc);
 48         else res2=tmp0;
 49     node res={0};
 50     res.smm=res1.smm+res2.smm;
 51     res.mx=max(res1.mx,res2.mx);
 52     res.mx=max(res.mx,res1.mr+res2.ml);
 53     res.ml=max(res1.ml,res1.smm+res2.ml);
 54     res.mr=max(res2.mr,res2.smm+res1.mr);
 55     return res;
 56 }
 57 int qsum(int L,int R,int l,int r,int rt){
 58     if(L<=l && r<=R)return t[rt].smm;
 59     int mid=(l+r)>>1;
 60     int res=0;
 61     if(L<=mid)res+=qsum(L,R,l,mid,lc);
 62     if(R>mid)res+=qsum(L,R,mid+1,r,rc);
 63     return res;
 64 }
 65 int main(){
 66     int T;
 67     T=read();
 68     while(T--){
 69         n=read();
 70         int i,j,x0,y0,x2,y2;
 71         for(i=1;i<=n;i++)data[i]=read();
 72         Build(1,n,1);
 73         m=read();
 74         tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=0;
 75         for(i=1;i<=m;i++){
 76             x0=read();y0=read();x2=read();y2=read();
 77             int tmp=0;
 78             int ans=-1e9;
 79             if(y0>=x2){
 80             //区间重叠
 81                 node res=query(x2,y0,1,n,1);
 82                 node res1=query(x0,x2-1,1,n,1);
 83                 node res2=query(y0+1,y2,1,n,1);
 84                 ans=max(ans,res1.mr+res.smm+res2.ml);
 85                 ans=max(ans,res1.mr+res.ml);
 86                 ans=max(ans,res.mr+res2.ml);
 87                 ans=max(ans,res.mx);
 88             }
 89             else{
 90             //区间未重叠
 91                 if(y0+1<x2)tmp=qsum(y0+1,x2-1,1,n,1);
 92                 node res1=query(x0,y0,1,n,1);
 93                 node res2=query(x2,y2,1,n,1);
 94                 ans=max(ans,tmp+res1.mr+res2.ml);
 95             }
 96             //printf("%d\n",query(x2,y2,1,n,1).mx);
 97             printf("%d\n",ans);
 98         }
 99     }
100     return 0;
101 }