제출 #225972

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
225972		balbit	Dragon 2 (JOI17_dragon2)	C++14	100 / 100	3744 ms	7544 KiB

dragon2

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

#define ll long long
#define pii pair<int,int>
#define f first
#define s second
#define SZ(x) (int)(x.size())
#define ALL(x) x.begin(),x.end()
#define pb push_back

#ifdef BALBIT
#define bug(...) cerr<<__LINE__<<": "<<#__VA_ARGS__<<": ", _do(__VA_ARGS__)
template<typename T> void _do(T && x){cerr<<x<<endl;}
template<typename T, typename ...S> void _do(T && x, S&&...y){cerr<<x<<", "; _do(y...);}
#define IOS()
#else
#define IOS() ios::sync_with_stdio(0),cin.tie(0)
#define endl '\n'
#define bug(...)
#endif // BALBIT

#define double long double
#define acos acosl
#define hypot hypotl

const int maxn = 1e5+5;

struct pt{
    ll x, y;
    pt operator - (pt oth) {
    return {x-oth.x, y-oth.y};
    }
    ll operator * (pt oth) {
        return x * (ll) oth.x + y * (ll) oth.y;
    }
    ll operator ^ (pt oth) {
        return x * (ll) oth.y - y * (ll) oth.x;
    }
};

struct Dg{
    pt p;
    double A, B;
    bool above;
};

double getang(pt v, pt u) {
    // from v to u, clockwise
    return ((v*u) / (hypot(v.x, v.y) * hypot(u.x, u.y)));
}
const double PI = acos(-1);
vector<Dg> G[maxn];
signed main(){
    int n,m; cin>>n>>m;
    for(int i = 0; i<n; ++i) {
        int a,b,c; cin>>a>>b>>c;
        G[c].pb({{a,b},0,0});
    }
    pt S, T; cin>>S.x>>S.y>>T.x>>T.y;
    for (int i = 1; i<=m; ++i) {
        for (Dg & d : G[i]) {
            d.A = getang(d.p-S, T-S);
            d.B = getang(d.p-T, S-T);
            bug(d.p.x, d.p.y);
            bug(d.A, d.B);
            d.above = ((T-S) ^ (d.p - S)) > 0;
            bug(d.above);
        }
    }
    int Q; cin>>Q;
    while (Q--) {
        int a,b; cin>>a>>b;
        int ret = 0;
        for (Dg &d1 : G[a]) {
            for (Dg & d2 : G[b]) {
                if (d1.above == d2.above) {
                    ret += ((((d1.p-S)^(d2.p-S))>0) != ((((d1.p-T)^(d2.p-T))) > 0)) && ((((T-S) ^ (d2.p-S)) > 0) == (((d2.p-S) ^ (d1.p-S)) > 0));
                }else{
                    ret +=  ((((d1.p-S)^(d2.p-S))>0) != ((((d1.p-T)^(d2.p-T))) > 0));
                }
            }
        }
        cout<<ret<<endl;
    }
}

• Subtask 1: 15 / 15
• Subtask 2: 45 / 45
• Subtask 3: 40 / 40