답안 #69442

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
69442 2018-08-20T21:08:53 Z Benq Dragon 2 (JOI17_dragon2) C++14
0 / 100
100 ms 8680 KB
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
 
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
 
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
 
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
 
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 30001;
 
const ld PI = 4*atan((ld)1);
 
namespace geo {
    istream& operator>> (istream& is, pi& p) {
        is >> p.f >> p.s;
        return is;
    }
    void nor(pd& x) { if (x.f < -PI) x.f += 2*PI, x.s += 2*PI; }
    ld area(cd a, cd b, cd c) { return (conj(b-a)*(c-a)).imag(); }
    pi operator-(const pi& l, const pi& r) { return {l.f-r.f,l.s-r.s}; }
}

using namespace geo;

struct BIT {
    vector<array<int,3>> toUpd;
    vector<pair<array<int,3>,int*>> toQuery;
    vi m, bit;
    
    void clr() {
        toUpd.clear(), toQuery.clear(), m.clear(), bit.clear();
    }
    
    void upd(int x, int y) {
        for (int X = ub(all(m),x)-m.begin()-1; X < sz(bit); X += (X&-X))
            bit[X] += y;
    }
    
    void query(int x, int y, int* z) {
        for (int X = ub(all(m),x)-m.begin()-1; X; X -= (X&-X)) 
            (*z) += y*bit[X];
    }
    
    void prop() {
        for (auto x: toUpd) m.pb(x[1]);
        m.pb(-MOD); sort(all(m)); m.erase(unique(all(m)),m.end());
        bit.resize(sz(m));
        sort(all(toUpd)), sort(all(toQuery));
        
        int ind = 0;
        for (auto x: toQuery) {
            while (ind < sz(toUpd) && toUpd[ind][0] <= x.f[0]) {
                upd(toUpd[ind][1],toUpd[ind][2]);
                ind ++;
            } 
            query(x.f[1],x.f[2],x.s);
        }
    }
};

int N,M,group[MX],ans[100001];
vi member[MX];
pi pos[MX], h[2];
pi POS[MX];
pair<pi,pi> BOUND[MX];
vpi query[MX], query2[MX];
BIT z;

void process1(int x) {
    z.clr();
    array<int,MX> co = array<int,MX>(); 
    for (int a: member[x]) {
        z.toUpd.pb({BOUND[a].f.f+1,BOUND[a].s.f+1,1});
        z.toUpd.pb({BOUND[a].f.s+1,BOUND[a].s.s+1,1});
        z.toUpd.pb({BOUND[a].f.f+1,BOUND[a].s.s+1,-1});
        z.toUpd.pb({BOUND[a].f.s+1,BOUND[a].s.f+1,-1});
    }
    FOR(a,1,N+1) {
        z.toQuery.pb({{POS[a].f,POS[a].s,1},&co[group[a]]});
        z.toQuery.pb({{POS[a].f+N,POS[a].s,1},&co[group[a]]});
        z.toQuery.pb({{POS[a].f,POS[a].s+N,1},&co[group[a]]});
        z.toQuery.pb({{POS[a].f+N,POS[a].s+N,1},&co[group[a]]});
    }
    z.prop(); 
    for (auto a: query[x]) ans[a.s] = co[a.f]; 
}

void process2(int x) {
    for (auto a: query[x]) query2[a.f].pb({x,a.s}); 
}

void process(int x) {
    //if (sz(query[x])) process1(x);
    if ((ll)sz(member[x])*sz(query[x]) >= N) process1(x);
    else process2(x);
}

int half(pi x) {
    if (x.s != 0) return x.s > 0;
    return x.f > 0;
}

ll area(pi a, pi b) {
    return (ll)a.f*b.s-(ll)a.s*b.f;
}

ll area(pi a, pi b, pi c) {
    b.f -= a.f, b.s -= a.s;
    c.f -= a.f, c.s -= a.s;
    return area(b,c);
}

bool cmp(pi a, pi b) {
    if (half(a) != half(b)) return half(a) < half(b);
    return area(a,b) > 0;
}

pi nor(pi x) {
    while (x.f < 0) x.f += N, x.s += N;
    while (x.f >= N) x.f -= N, x.s -= N;
    /*if (x.s < 0 || x.s > 2*N-1 || x.s <= x.f || x.s > x.f+N) {
        cerr << " " << x.f << " " << x.s << "\n";
        exit(0);
    }*/
    return x;
}

void genCoordinate(int ind) {
    vector<pair<pi,int>> v;
    FOR(i,1,N+1) v.pb({pos[i]-h[ind],i});
    sort(all(v),[](auto a, auto b) { return cmp(a.f,b.f); });
    int cur = 0;
    F0R(i,N) {
        while (cur < i+N && area(v[i].f,v[cur%N].f) >= 0) cur ++;
        if (ind == 0) POS[v[i].s].f = i;
        else POS[v[i].s].s = i;
        
        if (area(h[0],h[1],pos[v[i].s]) > 0) {
            if (ind == 0) BOUND[v[i].s].f = nor({cur-1,i+N-1});
            else BOUND[v[i].s].s = nor({i,cur-1});
        } else {
            if (ind == 0) BOUND[v[i].s].f = nor({i,cur-1});
            else BOUND[v[i].s].s = nor({cur-1,i+N-1});
        }
    }
}

void input() {
    //freopen("Input.txt","r",stdin);
    //freopen("Output.txt","w",stdout);
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M;
    FOR(i,1,N+1) {
        cin >> pos[i] >> group[i];
        member[group[i]].pb(i);
    }
    cin >> h[0] >> h[1];
    genCoordinate(0);
    genCoordinate(1);
}
 
int main() {
    input();
    int Q; cin >> Q; Q = 50;
    F0R(i,Q) {
        int f,g; cin >> f >> g;
        query[f].pb({g,i});
    }
    FOR(i,1,M+1) process(i);
    FOR(i,1,M+1) {
        z.clr();
        for (auto a: query2[i]) for (int b: member[a.f]) {
            assert(group[b] != i);
            z.toQuery.pb({{BOUND[b].f.s,BOUND[b].s.s,1},&ans[a.s]});
            z.toQuery.pb({{BOUND[b].f.f,BOUND[b].s.s,-1},&ans[a.s]});
            z.toQuery.pb({{BOUND[b].f.s,BOUND[b].s.f,-1},&ans[a.s]});
            z.toQuery.pb({{BOUND[b].f.f,BOUND[b].s.f,1},&ans[a.s]});
        }
        for (int a: member[i]) {
            z.toUpd.pb({POS[a].f,POS[a].s,1});
            z.toUpd.pb({POS[a].f+N,POS[a].s,1});
            z.toUpd.pb({POS[a].f,POS[a].s+N,1});
            z.toUpd.pb({POS[a].f+N,POS[a].s+N,1});
        }
        z.prop();
    }
    F0R(i,Q) cout << ans[i] << "\n";
}
 
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/
# 결과 실행 시간 메모리 Grader output
1 Incorrect 15 ms 3448 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 100 ms 8680 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 15 ms 3448 KB Output isn't correct
2 Halted 0 ms 0 KB -