Problem Link : http://www.spoj.com/problems/GSS2/
Solution Idea:
-
Hint : https://www.quora.com/How-can-the-SPOJ-problem-GSS2-be-solved/answer/Brian-Bi?srid=29u8
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pii pair <int,int>
#define pll pair <long long,long long>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define db double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define sqr(x) (x)*(x)
#define D(x) cerr<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(i,a,b) for(int i=a;i>=b;i--)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d: ",z)
#define LINE_PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*------------------------------------------------*/
/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/
#define mx 100005
#define segment_tree int l=n*2,r=l+1,mid=(b+e)/2
int ara[mx+5];
int last[2*mx+5];
ll ans[mx+5];
struct data
{
int l,r,id;
};
vector<data>queries;
struct data1
{
ll sum,lazy,best_lazy,ans;
data1()
{
sum=lazy=best_lazy=ans=0;
}
};
data1 tree[3*mx];
void push_down(data1 &cur, data1 &lft, data1 &rgt)
{
// if(cur.lazy || cur.best_lazy)
// {
lft.best_lazy=max(lft.best_lazy,lft.lazy+cur.best_lazy);
lft.lazy+=cur.lazy;
lft.ans=max(lft.ans,lft.sum+cur.best_lazy);
lft.sum=lft.sum+cur.lazy;
rgt.best_lazy=max(rgt.best_lazy,rgt.lazy+cur.best_lazy);
rgt.lazy+=cur.lazy;
rgt.ans=max(rgt.ans,rgt.sum+cur.best_lazy);
rgt.sum+=cur.lazy;
cur.lazy=cur.best_lazy=0;
// }
}
void push_up(data1 &cur, data1 &lft, data1 &rgt)
{
cur.ans=max(lft.ans,rgt.ans);
cur.sum=max(lft.sum,rgt.sum);
}
void update(int n, int b, int e, int i, int j, ll val)
{
if(b>j || e<i) return;
if(b>=i && e<=j)
{
tree[n].best_lazy=max(tree[n].best_lazy,tree[n].lazy+val);
tree[n].lazy+=val;
tree[n].ans=max(tree[n].ans,tree[n].sum+val);
tree[n].sum+=val;
return;
}
segment_tree;
push_down(tree[n],tree[l],tree[r]);
update(l,b,mid,i,j,val);
update(r,mid+1,e,i,j,val);
push_up(tree[n],tree[l],tree[r]);
}
data1 query(int n, int b, int e, int i, int j)
{
if(b>j || e<i) return data1();
if(b>=i && e<=j)
{
return tree[n];
}
segment_tree;
push_down(tree[n],tree[l],tree[r]);
data1 p=query(l,b,mid,i,j);
data1 q=query(r,mid+1,e,i,j);
data1 ret;
push_up(ret,p,q);
return ret;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n;
sf(n);
for(int i=1; i<=n; i++) sf(ara[i]);
int m;
sf(m);
data a;
for(int i=0; i<m; i++)
{
sff(a.l,a.r);
a.id=i;
queries.pb(a);
}
sort(all(queries),[](data a,data b)
{
return a.r<b.r;
});
int idx=0;
for(int i=1; i<=n && idx<SZ(queries); i++)
{
// if(last[mx+ara[i]]!=0)
// {
// update(1,1,n,last[mx+ara[i]],0);
// }
int xx=last[mx+ara[i]]+1;
update(1,1,n,xx,i,ara[i]);
last[mx+ara[i]]=i;
while(idx<SZ(queries) && queries[idx].r==i)
{
ans[queries[idx].id]=query(1,1,n,queries[idx].l,queries[idx].r).ans;
idx++;
}
}
for(int i=0; i<m; i++)
printf("%lld\n",ans[i]);
return 0;
}