Submission #893808

#	Time	Username	Problem	Language	Result	Execution time	Memory
893808		yfrank792	Sjeckanje (COCI21_sjeckanje)	C++14	110 / 110	487 ms	36436 KiB

Main

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef queue<int> qi;
typedef set<int> si;

typedef priority_queue<int> pqi;
typedef priority_queue<ll> pql;
typedef priority_queue<int,vi,greater<int> > pqi_g;
typedef priority_queue<ll,vll,greater<ll> > pql_g;

#define F first
#define S second
#define PB push_back
#define PF push_front
#define MP make_pair
#define FOR(i,a,b) for (int i=a; i<=b; i++)
#define _FOR(i,a,b) for (int i=a; i>=b; i--)
#define F0R(i,a,b,x) for (int i=a; i<=b; i+=x)
#define FOR_(x,v) for (auto x : v)
#define F0R_(x,v) for (auto &x : v)

const int INF = 0x3f3f3f3f;
const int maxn = 2e5+2;
const ll MOD = 1e9+7;

//? link: https://oj.uz/problem/view/COCI21_sjeckanje

//? sol: segtree
//          - it can be observed that the segments that will produce the max total value are always either increasing or decreasing
//          - since the max total value is sum of the difference b/t the max and min values in each segment, we can store the array as one of differences
//          - this way, the value for a segment is just the sum of the differences in that segment
//          - when making updates, we then only need to change the end values of the segments that are affected
//          - we can then use a segtree to store the difference array
//          - since each segment must be one spaced apart (can't use the same element twice), we store the max value for each segtree node given 4 scenarios:
//              - take the left value and the right value
//              - take the left value and not the right value
//              - take the right value and not the left value
//              - take neither the left value nor the right value
//          - we also store the left and right border values for each node

int n, q;
class segtree {
    struct node {ll yy,yn,ny,nn,l,r;} d[4*maxn];
    private:
        ll inline ls(ll x) {return x<<1;}
        ll inline rs(ll x) {return x<<1|1;}
        node inline op(node &l,node &r) {
            node ret = {
                max(l.yn+r.yy, l.yy+r.ny),          // yy
                max(l.yy+r.nn, l.yn+r.yn),          // yn
                max(l.ny+r.ny, l.nn+r.yy),          // ny
                max(l.ny+r.nn, l.nn+r.yn),          // nn
                l.l,                                // l
                r.r                                 // r
            };
            if ((l.r>=0 && r.l>=0) || (l.r<=0 && r.l<=0)) {
                ret.yy = max(ret.yy, l.yy+r.yy);
                ret.yn = max(ret.yn, l.yy+r.yn);
                ret.ny = max(ret.ny, l.ny+r.yy);
                ret.nn = max(ret.nn, l.ny+r.yn);
            }
            return ret;
        }
    public:
        ll a[maxn];
        void init(ll s=1,ll t=n,ll p=1) {
            if (s==t) {
                d[p] = {abs(a[s]),0,0,0,a[s],a[s]};
                return;
            }
            ll mid = s + ((t-s)>>1);
            init(s, mid, ls(p));
            init(mid+1, t, rs(p));
            d[p] = op(d[ls(p)], d[rs(p)]);
        }
        void upd(ll v,ll x,ll s=1,ll t=n,ll p=1) {      // upd pos v by value x ~ p=cur pos, s,t=cur range
            if (s==t) {
                a[s] += x;
                d[p] = {abs(a[s]),0,0,0,a[s],a[s]};
                return;
            }
            ll mid = s + ((t-s)>>1);
            if (v<=mid) upd(v,x,s,mid,ls(p));
            else upd(v,x,mid+1,t,rs(p));
            d[p] = op(d[ls(p)], d[rs(p)]);
        }
        ll query() {
            return max(d[1].yy, max(d[1].yn, max(d[1].ny, d[1].nn)));
        }
        void output_a() {
            FOR(i,1,n) cout << a[i] << " ";
            cout << endl;
        }
} st;

int main()
{
    // faster cin/cout
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    // freopen("c.in","r",stdin);

    cin >> n >> q;
    int last; cin>>last; st.a[1]=0;
    FOR(i,2,n) {
        int x; cin>>x;
        st.a[i] = x-last;
        last = x;
    }
    st.init();

    FOR(_,1,q) {
        int l,r,x; cin>>l>>r>>x;
        if (l!=1) st.upd(l,x);
        if (r!=n) st.upd(r+1,-x);
        // st.output_a();
        cout << st.query() << endl;
    }

    return 0;   
}

• Subtask 1: 15 / 15
• Subtask 2: 40 / 40
• Subtask 3: 55 / 55