Submission #216704

#	Time	Username	Problem	Language	Result	Execution time	Memory
216704		combi1k1	Sweeping (JOI20_sweeping)	C++14	0 / 100	1329 ms	307820 KiB

sweeping

//this solution is referred from comment of 300iq:
//https://codeforces.com/blog/entry/74871?#comment-590823
#include<bits/stdc++.h>

using namespace std;

#define ll  long long
#define ld  double

#define sz(x)   (int)x.size()
#define all(x)  x.begin(),x.end()

#define pb  emplace_back
#define X   first
#define Y   second

const int   N   = 2e6 + 5;

typedef pair<int,int>   ii;
typedef pair<int,ii>    pii;

struct Node {
    Node *l, *r;
    Node *par;
    ii  key;
    int pri, nCh;
    int lzX, lzY;

    Node(int x = 0,int y = 0) : l(0), r(0), par(0), key(ii(x,y)), pri(rand() << 16 ^ rand()), nCh(1), lzX(-1), lzY(-1)  {}
}  *p[N];

int Siz(Node *x)    {
    if (x)  return  x -> nCh;
    else    return  0;
}

void app(Node *x,int a,int b)   {
    if (a >= 0) (x -> key).X = x -> lzX = a;
    if (b >= 0) (x -> key).Y = x -> lzY = b;
}
void pull(Node *x)  {
    x -> nCh = 1;
    x -> nCh += Siz(x -> l);
    x -> nCh += Siz(x -> r);

    if (x -> l) x -> l -> par = x;
    if (x -> r) x -> r -> par = x;
}
void push(Node *x)  {
    if (x -> l) app(x -> l,x -> lzX,x -> lzY);
    if (x -> r) app(x -> r,x -> lzX,x -> lzY);

    x -> lzX = -1;
    x -> lzY = -1;
}
Node *join(Node *x,Node *y) {
    if (!x) return  y;
    if (!y) return  x;

    if (x -> pri > y -> pri)    {
        push(x);    x -> r = join(x -> r,y);
        pull(x);    return  x;
    }
    else    {
        push(y);    y -> l = join(x,y -> l);
        pull(y);    return  y;
    }
}
void split(Node *x,Node*&l,Node*&r,ii  v)   {
    if (!x) {
        l = r = 0;
        return;
    }
    push(x);

    if (v.X < (x -> key).X) {   split(x -> l,l,x -> l,v);   pull(r = x);    return; }
    if (v.Y < (x -> key).Y) {   split(x -> l,l,x -> l,v);   pull(r = x);    return; }

    split(x -> r,x -> r,r,v);
    pull(l = x);
}

vector<ii>  vec;

void dfs(Node *x)   {
    if(!x)  return;

    dfs(x -> l);    vec.pb(x -> key);
    dfs(x -> r);
}

struct Group    {
    Node *Rt;
    int x, y;
};
bool operator < (const Group&a,const Group&b)   {
    ii  tmp1(-a.y,a.Rt -> pri);
    ii  tmp2(-b.y,b.Rt -> pri);

    return  tmp1 < tmp2;
}

priority_queue<Group>   pq[N];

int tr[N << 2];

#define lch     i << 1
#define rch     i << 1 | 1

void upd(int i,int l,int r,int p)   {
    if (l > p)  return;
    if (r < p)  return;
    if (l == r) {
        if (pq[l].size())   tr[i] = pq[l].top().y;
        else                tr[i] = 1e9 + 7;
        return;
    }
    int m = (l + r) / 2;

    upd(lch,l,m,p); ++m;
    upd(rch,m,r,p);

    tr[i] = min(tr[lch],tr[rch]);
}

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    srand(time(NULL));

    int n;  cin >> n;
    int m;  cin >> m;
    int q;  cin >> q;

    vector<pii> Que;
    vector<int> val;

    for(int i = 0 ; i < m ; ++i)    {
        int x;  cin >> x;
        int y;  cin >> y;

        Que.pb(4,ii(x,y));
        val.pb(x);
    }
    for(int i = 0 ; i < q ; ++i)    {
        int t;  cin >> t;
        int x;  cin >> x;
        int y = n - x;

        if (t == 4) cin >> y;
        if (t == 2) swap(x,y);

        Que.pb(t,ii(x,y));

        if (t != 1)
            val.pb(x);
    }
    sort(all(val));
    val.erase(unique(all(val)),val.end());

    auto add = [&](Node *R) {
        if(!R)  return;
        Group G;
        G.Rt = R;

        while (R -> l)
            push(R),
            R = R -> l;

        G.x = (R -> key).X;
        G.y = (R -> key).Y;

        int p = upper_bound(all(val),G.x) - val.begin();

        pq[p].push(G);
        upd(1,1,sz(val),p);
    };

    fill(tr + 1,tr + 4 * sz(val),1e9 + 7);
    int tot = 0;

    for(pii T : Que)    {
        int t = T.X;
        int x = T.Y.X;
        int y = T.Y.Y;

        if (t == 1) {
            vector<Node*> Path;

            for(Node* a = p[x] ; a -> par ; Path.pb(a = a -> par)); reverse(all(Path));
            for(Node* a : Path) push(a);

            cout << (p[x] -> key).X << " ";
            cout << (p[x] -> key).Y << "\n";
            continue;
        }
        if (t == 4) {
            add(p[++tot] = new Node(x,y));
            continue;
        }
        vector<Node*> tmp;

        while (tr[1] <= y)  {
            int i = 1;
            int l = 1;
            int r = sz(val);

            while (l < r)   {
                int m = (l + r) / 2;

                if (tr[lch] <= y)   i = lch, r = m;
                else                i = rch, l = m + 1;
            }
            if (val[l - 1] > x)
                break;

            while (pq[l].size())    {
                if(pq[l].top().y > y)
                    break;

                Group G = pq[l].top();  pq[l].pop();

                Node *a;
                Node *b;

                split(G.Rt,a,b,ii(x,y));    add(b);

                if (a)  a -> par = 0;
                if (b)  b -> par = 0;

                assert(a);
                tmp.pb(a);
            }
            upd(1,1,sz(val),l);
        }
        sort(all(tmp),[&](Node *a,Node *b)  {
            return  Siz(a) > Siz(b);
        });
        for(Node *a : tmp)  {
            if (t == 2) app(a,x,-1);
            if (t == 3) app(a,-1,y);
        }
        Node *Rt = 0;

        for(Node *a : tmp)  {
            if (Siz(Rt) < Siz(a))
                swap(Rt,a);

            vec.clear();
            dfs(a);

            for(ii  e : vec)    {
                Node *x;
                Node *y;

                split(a,x,a,e);     assert(x && x -> key == e);
                split(Rt,Rt,y,e);

                Rt = join(Rt,x);
                Rt = join(Rt,y);
            }
        }
        add(Rt);
    }
}

Compilation message (stderr)

sweeping.cpp: In function 'int main()':
sweeping.cpp:191:13: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
             for(Node* a = p[x] ; a -> par ; Path.pb(a = a -> par)); reverse(all(Path));
             ^~~
sweeping.cpp:191:69: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
             for(Node* a = p[x] ; a -> par ; Path.pb(a = a -> par)); reverse(all(Path));
                                                                     ^~~~~~~

• Subtask 1: 0 / 1
• Subtask 2: 0 / 10
• Subtask 3: 0 / 11
• Subtask 4: 0 / 53
• Subtask 5: 0 / 25