oj.uz

답안 #69407

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
69407 2018-08-20T19:08:43 Z Benq Dragon 2 (JOI17_dragon2) C++14
15 / 100
 4000 ms 19996 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;
 
namespace geo {
    template<class T> istream& operator>> (istream& is, complex<T>& p) {
        T value;
        is >> value; p.real(value);
        is >> value; p.imag(value);
        return is;
    }
    void nor(pd& x) { if (x.f < -M_PIl) x.f += 2*M_PIl, x.s += 2*M_PIl; }
    ld area(cd a, cd b, cd c) { return (conj(b-a)*(c-a)).imag(); }
}
 
using namespace geo;
 
struct BIT {
    vector<array<ld,3>> toUpd;
    vector<pair<array<ld,3>,int*>> toQuery;
    vd m;
    vi bit;
    
    void clr() {
        toUpd.clear(), toQuery.clear(), m.clear(), bit.clear();
    }
    
    void upd(ld x, int y) {
        for (int X = ub(all(m),x)-m.begin()-1; X < sz(bit); X += (X&-X))
            bit[X] += y;
    }
    
    void query(ld 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];
cd pos[MX], h[2];
pd POS[MX];
pair<pd,pd> BOUND[MX];
vpi query[MX], query2[MX];
BIT z;
 
void process1(int x) {
    z.clr();
    array<int,MX> co; co.fill(0);
    for (int a: member[x]) {
        z.toUpd.pb({BOUND[a].f.f,BOUND[a].s.f,1});
        z.toUpd.pb({BOUND[a].f.s,BOUND[a].s.s,1});
        z.toUpd.pb({BOUND[a].f.f,BOUND[a].s.s,-1});
        z.toUpd.pb({BOUND[a].f.s,BOUND[a].s.f,-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+2*M_PIl,POS[a].s,1},&co[group[a]]});
        z.toQuery.pb({{POS[a].f,POS[a].s+2*M_PIl,1},&co[group[a]]});
        z.toQuery.pb({{POS[a].f+2*M_PIl,POS[a].s+2*M_PIl,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(member[x]) >= 150) process1(x);
    else process2(x);
}
 
int get() {
    int z = 1e9;
    return rand() % (2*z+1)-z;
}
 
cd gen() { return {get(),get()}; }
 
void input() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M;
    FOR(i,1,N+1) {
        //pos[i] = gen();
        //group[i] = rand() % M+1;
        cin >> pos[i] >> group[i];
        member[group[i]].pb(i);
    }
    //h[0] = gen();
    //h[1] = gen();
    F0R(i,2) cin >> h[i];
    FOR(i,1,N+1) {
        POS[i].f = arg(pos[i]-h[0]);
        POS[i].s = arg(pos[i]-h[1]);
        if (area(h[0],h[1],pos[i]) > 0) BOUND[i] = {{POS[i].f-M_PIl,POS[i].f},{POS[i].s,POS[i].s+M_PIl}};
        else BOUND[i] = {{POS[i].f,POS[i].f+M_PIl},{POS[i].s-M_PIl,POS[i].s}};
        nor(BOUND[i].f); nor(BOUND[i].s);
    }
}
 
int main() {
    input();
    int Q; cin >> Q;
    F0R(i,Q) {
        int f,g; cin >> f >> g;
        // int f = rand() % M+1, g = rand() % M+1;
        query[f].pb({g,i});
    }
    FOR(i,1,M+1) process(i);
    int tmp = 0;
    FOR(i,1,M+1) {
        z.clr();
        for (auto a: query2[i]) for (int b: member[a.f]) {
            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+2*M_PIl,POS[a].s,1});
            z.toUpd.pb({POS[a].f,POS[a].s+2*M_PIl,1});
            z.toUpd.pb({POS[a].f+2*M_PIl,POS[a].s+2*M_PIl,1});
        }
        tmp += sz(z.toQuery);
        z.prop();
    }
    // cout << tmp << "\n";
    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 :/
*/

Compilation message

dragon2.cpp: In function 'cd gen()':
dragon2.cpp:139:23: warning: narrowing conversion of 'get()' from 'int' to 'long double' inside { } [-Wnarrowing]
 cd gen() { return {get(),get()}; }
                    ~~~^~
dragon2.cpp:139:29: warning: narrowing conversion of 'get()' from 'int' to 'long double' inside { } [-Wnarrowing]
 cd gen() { return {get(),get()}; }
                          ~~~^~

Subtask #1
15.0 / 15.0

# 결과 실행 시간 메모리 Grader output
1 Correct 29 ms 4592 KB Output is correct
2 Correct 58 ms 4592 KB Output is correct
3 Correct 185 ms 4592 KB Output is correct
4 Correct 330 ms 6036 KB Output is correct
5 Correct 123 ms 6412 KB Output is correct
6 Correct 9 ms 6412 KB Output is correct
7 Correct 10 ms 6412 KB Output is correct
8 Correct 30 ms 6412 KB Output is correct
9 Correct 21 ms 6412 KB Output is correct
10 Correct 23 ms 6412 KB Output is correct

Subtask #2
0 / 45.0

# 결과 실행 시간 메모리 Grader output
1 Correct 266 ms 19996 KB Output is correct
2 Correct 624 ms 19996 KB Output is correct
3 Execution timed out 4027 ms 19996 KB Time limit exceeded
4 Halted 0 ms 0 KB -

Subtask #3
0 / 40.0

# 결과 실행 시간 메모리 Grader output
1 Correct 29 ms 4592 KB Output is correct
2 Correct 58 ms 4592 KB Output is correct
3 Correct 185 ms 4592 KB Output is correct
4 Correct 330 ms 6036 KB Output is correct
5 Correct 123 ms 6412 KB Output is correct
6 Correct 9 ms 6412 KB Output is correct
7 Correct 10 ms 6412 KB Output is correct
8 Correct 30 ms 6412 KB Output is correct
9 Correct 21 ms 6412 KB Output is correct
10 Correct 23 ms 6412 KB Output is correct
11 Correct 266 ms 19996 KB Output is correct
12 Correct 624 ms 19996 KB Output is correct
13 Execution timed out 4027 ms 19996 KB Time limit exceeded
14 Halted 0 ms 0 KB -