[SPOJ1557] Can you answer these queries II

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=2482

[算法]

线段树维护历史最值

时间复杂度 : O(NlogN)

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 200010
typedef long long ll;
typedef long double ld;
const int T = ;

struct query
{
        int l , r;
        int id;
} q[MAXN];

int n , m;
int loc[MAXN << ] , pre[MAXN << ] , val[MAXN << ];
ll ans[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}

struct Segment_Tree
{
        struct Node
        {
                int l , r;
                ll sum , hsum;
                ll taga , tagb;
        } a[MAXN << ];
        inline void build(int index , int l , int r)
        {
                a[index].l = l , a[index].r = r;
                if (l == r) return;
                int mid = (l + r) >> ;
                build(index <<  , l , mid);
                build(index <<  |  , mid +  , r);
        }
        inline void pushdown(int index)
        {
                int l = a[index].l , r = a[index].r;
                int mid = (l + r) >> ;
                if (l == r) return;
                a[index << ].hsum = max(a[index << ].hsum , a[index << ].sum + a[index].tagb);
                a[index <<  | ].hsum = max(a[index <<  | ].hsum , a[index <<  | ].sum + a[index].tagb);
                a[index << ].sum += a[index].taga;
                a[index <<  | ].sum += a[index].taga;
                chkmax(a[index << ].tagb , a[index << ].taga + a[index].tagb);
                chkmax(a[index <<  | ].tagb , a[index <<  | ].taga + a[index].tagb);
                a[index << ].taga += a[index].taga;
                a[index <<  | ].taga += a[index].taga;
                a[index].taga = a[index].tagb = ;
        }
        inline void update(int index)
        {
                a[index].sum = max(a[index << ].sum , a[index <<  | ].sum);
                a[index].hsum = max(a[index << ].hsum , a[index <<  | ].hsum);
        }
        inline void modify(int index , int l , int r , ll val)
        {
                pushdown(index);
                if (a[index].l == l && a[index].r == r)
                {
                        a[index].sum += val;
                        chkmax(a[index].hsum , a[index].sum);
                        a[index].taga += val;
                        chkmax(a[index].tagb , a[index].taga);
                } else
                {
                        int mid = (a[index].l + a[index].r) >> ;
                        if (mid >= r) modify(index <<  , l , r , val);
                        else if (mid +  <= l) modify(index <<  |  , l , r , val);
                        else
                        {
                                modify(index <<  , l , mid , val);
                                modify(index <<  |  , mid +  , r , val);
                        }
                        update(index);
                }
         }
        inline ll query(int index , int l , int r)
        {
                pushdown(index);
                if (a[index].l == l && a[index].r == r)
                        return a[index].hsum;
                int mid = (a[index].l + a[index].r) >> ;
                if (mid >= r) return query(index <<  , l , r);
                else if (mid +  <= l) return query(index <<  |  , l , r);
                else return max(query(index <<  , l , mid) , query(index <<  |  , mid +  , r));
        }
} SGT;

inline bool cmp(query a , query b)
{
        return a.r < b.r;
}

int main()
{

        read(n);
        for (int i = ; i <= n; i++) read(val[i]);
        read(m);
        for (int i = ; i <= m; i++)
        {
                read(q[i].l);
                read(q[i].r);
                q[i].id = i;
        }
        sort(q +  , q + m +  , cmp);
        for (int i = ; i <= n; i++)
        {
                pre[i] = loc[val[i] + T];
                loc[val[i] + T] = i;
        }
        SGT.build( ,  , n);
        int now = ;
        for (int i = ; i <= n; i++)
        {
                SGT.modify( , pre[i] +  , i , val[i]);
                while (now <= m && q[now].r == i)
                {
                        ans[q[now].id] = max(SGT.query( , q[now].l , q[now].r) , 0LL);
                        ++now;
                }
        }
        for (int i = ; i <= m; i++) printf("%lld\n" , ans[i]);

        return ;

}