SPOJ GSS2 / 洛谷 SP1557 | GSS2 - Can you answer these queries II

题目链接：

主要实现

维护一个 prepre 数组，preipre_i 表示与 aia_i 相同的上一个元素的位置。

将询问的 rr 从小到大排序。遍历整个 aa 数组，每次将 [prei+1,i][pre_i+1, i] 添加 aia_i，然后处理所有 r≤ir \le i 的询问。对于每次询问，查询 [l,r][l, r] 即可。

线段树部分

存储 sumsum / hxsumhxsum / lazy_addlazy\_add / lazy_maxlazy\_max，分别代表 区间和 / 区间和历史最大值 / 懒标记 / 懒标记历史最大值。

Pushdown:

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hxsum[ls] = max(hxsum[ls], sum[ls]+lazy_max[u]);
hxsum[rs] = max(hxsum[rs], sum[rs]+lazy_max[u]);
sum[ls] += lazy_add[u];
sum[rs] += lazy_add[u];
lazy_max[ls] = max(lazy_max[ls], lazy_add[ls]+lazy_max[u]);
lazy_max[rs] = max(lazy_max[rs], lazy_add[rs]+lazy_max[u]);
lazy_add[ls] += lazy_add[u];
lazy_add[rs] += lazy_add[u];
lazy_add[u] = 0;
lazy_max[u] = 0;

Pushup:

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sum[u] = max(sum[ls], sum[rs]);
hxsum[u] = max(hxsum[ls], hxsum[rs]);

Query:

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Inrange: return hxsum[u];
!OutOfRange: return max(query(ls),query(rs));

Maketag:

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sum[u] += uk;
hxsum[u] = max(hxsum[u], sum[u]);
lazy_add[u] += uk;
lazy_max[u] = max(lazy_max[u], lazy_add[u]);

Modify:

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Inrange: maketag(u);
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/******************************
  @GavinCQTD / 2025-01-04 08:45:31
  "GSS2 - Can you answer these queries II" From Luogu
  # https://www.luogu.com.cn/problem/SP1557
  1000 ms / 1 MB
******************************/
/* We all make choices, but in the end our choices make us. */

// #pragma GCC optimize(1,2,3,"Ofast","inline")

#define NDEBUG

#include <iostream>
#include <algorithm>
#include <cassert>
#include <cstring>
#include <iomanip>
#include <cstdio>
#include <cmath>
using namespace std;
inline void pass(){return;}
inline void hold(){while(1);}
#define endl "\n"
#define mp(a,b) make_pair(a,b)
#define ct(a) while(!a.empty()) a.pop()
#define cmax(a,b) a=max(a,b)
#define cmin(a,b) a=min(a,b)
#define cdb cerr<<"Debug: "
#define cwn cerr<<"Warn: "
#define debug(x) cerr<<"In Line "<<__LINE__<<": "<<#x<<" = "<<x<<"\n"
#define sp(a) fixed<<setprecision(a)
#define likely(x) __builtin_expect(!!(x), 1)
#define unlikely(x) __builtin_expect(!!(x), 0)
#define ll long long
#define ull unsigned long long
#define ld long double
// #define int ll
// #define _debug

struct query{
    int l,r,id;
}qs[100005];

namespace ds{
    struct seg{
        struct segNode{
            int sum,hxsum,lzy,lzymax;
        }tree[400005];
        inline void pushup(int u){
            tree[u].sum = max(tree[u*2].sum, tree[u*2+1].sum);
            tree[u].hxsum = max(tree[u*2].hxsum, tree[u*2+1].hxsum);
        }
        inline bool inrange(int l,int r,int ql,int qr){return (ql<=l)&&(r<=qr);}
        inline bool outofrange(int l,int r,int ql,int qr){return (l>qr)||(r<ql);}
        inline void maketag(int u,int ul,int ur,int uk){
            tree[u].lzy += uk;
            cmax(tree[u].lzymax, tree[u].lzy);
            tree[u].sum += uk;
            cmax(tree[u].hxsum, tree[u].sum);
        }
        inline void pushdown(int u){
            cmax(tree[u*2].hxsum, tree[u*2].sum+tree[u].lzymax);
            cmax(tree[u*2+1].hxsum, tree[u*2+1].sum+tree[u].lzymax);
            tree[u*2].sum += tree[u].lzy;
            tree[u*2+1].sum += tree[u].lzy;
            cmax(tree[u*2].lzymax, tree[u*2].lzy+tree[u].lzymax);
            cmax(tree[u*2+1].lzymax, tree[u*2+1].lzy+tree[u].lzymax);
            tree[u*2].lzy += tree[u].lzy;
            tree[u*2+1].lzy += tree[u].lzy;
            tree[u].lzy = 0;
            tree[u].lzymax = 0;
        }
        int query(int u,int l,int r,int ql,int qr){
            if(inrange(l,r,ql,qr)) return tree[u].hxsum;
            else if(!outofrange(l,r,ql,qr)){
                int mid=(l+r)>>1;
                pushdown(u);
                return max(query(u*2,l,mid,ql,qr),query(u*2+1,mid+1,r,ql,qr));
            }
            else return 0;
        }
        void update(int u,int l,int r,int ul,int ur,int uk){
            if(inrange(l,r,ul,ur)) maketag(u,l,r,uk);
            else if(!outofrange(l,r,ul,ur)){
                int mid=(l+r)>>1;
                pushdown(u);
                update(u*2,l,mid,ul,ur,uk);
                update(u*2+1,mid+1,r,ul,ur,uk);
                pushup(u);
            }
        }
#ifndef NDEBUG
        inline void output(int u,int l,int r){
            cerr << "On node " << u << ", left " << l << ", right " << r << ", sum=" << tree[u].sum << ", hxsum="
                << tree[u].hxsum << ", lzy=" << tree[u].lzy << ", lzymax=" << tree[u].lzymax << ";\n";
            if(l==r) return;
            int mid=(l+r)>>1;
            output(u*2,l,mid);
            output(u*2+1,mid+1,r);
        }
#else
        inline void output(int u,int l,int r){}
#endif
    };
}using ds::seg;

bool cmp(query x,query y){
    return x.r<y.r;
}

#define OFFSET 100000
int _t,n,q,a[100005],pre[100005],curr[200005],ans[100005];
seg segobj;

void solve(int testID){
    cin >> n;
    for(int i=1;i<=n;i++){
        cin >> a[i];
        pre[i] = curr[a[i]+OFFSET];
        curr[a[i]+OFFSET] = i;
    }
    cin >> q;
    for(int i=1;i<=q;i++){
        cin >> qs[i].l >> qs[i].r;
        qs[i].id = i;
    }
    sort(qs+1,qs+q+1,cmp);
    // for(int i=1;i<=q;i++){
    //     segobj.update(1,1,n,pre[qs[i].r]+1,qs[i].r,a[qs[i].r]);
    //     ans[qs[i].id] = segobj.query(1,1,n,qs[i].l,qs[i].r);
    // }
    int j=1;
    for(int i=1;i<=n;i++){
        segobj.update(1,1,n,pre[i]+1,i,a[i]);
        for(;j<=q&&qs[j].r<=i;j++){
            ans[qs[j].id] = segobj.query(1,1,n,qs[j].l,qs[j].r);
        }
    }
    segobj.output(1,1,n);
    for(int i=1;i<=q;i++) cout<<ans[i]<<"\n";
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    // freopen(".in", "r", stdin);
    // freopen(".out", "w", stdout);
    // freopen("PublicDebug.debug", "w", stderr);

    // cout << sp();
    // cerr << sp();

    _t = 1;
    // cin >> _t;

    for(int ran=0;ran<_t;ran++) solve(ran);

    fflush(stdout);
    return 0;
}

/*
Main:

一个 pre 数组，pre[i] 表示与 a[i] 相同的上一个元素的位置

将询问的 r 从小到大排序。遍历整个 a 数组，每次将 [pre[i]+1, i] 添加 a[i]，然后处理所有 r <= i 的询问。对于每次询问，查询 [l, r] 即可。


Seg:

存储 sum / hxsum / lazy_add / lazy_max

Pushdown: ---------------------------------------
cmax(hxsum[ls], sum[ls]+lazy_max[u]);

sum[ls] += lazy_add[u];

cmax(lazy_max[ls], lazy_add[ls]+lazy_max[u]);

lazy_add[ls] += lazy_add[u];

lazy_add[u] = 0;
lazy_max[u] = 0;
-------------------------------------------------

Pushup: -----------------------------------------
sum[u] = max(sum[ls], sum[rs]);
hxsum[u] = max(hxsum[ls], hxsum[rs]);
-------------------------------------------------

Query: ------------------------------------------
Inrange: return hxsum[u];
!OutOfRange: return max(query(ls),query(rs));
-------------------------------------------------

Maketag: ----------------------------------------
sum[u] += uk;
cmax(hxsum[u], sum[u]);
lazy_add[u] += uk;
cmax(lazy_max[u], lazy_add[u]);
-------------------------------------------------

Modify: -----------------------------------------
Inrange: maketag(u);
-------------------------------------------------
*/
OI
#题解 #OI:线段树
SPOJ GSS2 / 洛谷 SP1557 | GSS2 - Can you answer these queries II
https://gib716-blog.netlify.app/2025/spoj-gss2-sp1557-gss2-can-you-answer-these-queries-ii/
作者
Gavin
发布于
2025年1月4日
许可协议