No.202 1円玉投げ

No.202 1円玉投げ - yukicoder

• バケット法を使う。分割領域に含まれる一円玉の最大個数はたかだか 25 個程度なので十分間に合う。

class ThrowOneYenCoin {
public:
    static const int r = 10;
    typedef pair<int,int> Pt;
    bool hit(const Pt &a, const Pt &b) {
            int ax,ay,bx,by;
            tie(ax,ay) = a;
            tie(bx,by) = b;
            return (ax-bx)*(ax-bx)+(ay-by)*(ay-by) < 4*r*r;
    }
    void solve(void) {
            int N;
            cin>>N;

            const int B = 100;
            const int maxX = 20000;
            const int maxY = 20000;
            const int maxBX = maxX/B;
            const int maxBY = maxY/B;
            list<Pt> field[maxBY+1][maxBX+1];

            // 100*100 に含まれるコインの数は最大25程度
            // O(N*9*25)
            REP(i,N)
            {
                int x,y;
                cin>>x>>y;

                Pt p(x,y);
                int bx = x/B;
                int by = y/B;

                int dx[] = {-1,0,1};
                int dy[] = {-1,0,1};

                bool ok = true;
                REP(j,3)
                REP(k,3)
                {
                    int nx = bx+dx[j];
                    int ny = by+dy[k];

                    if (nx < 0 || ny < 0 || maxBX < nx || maxBY < ny)
                        continue;

                    auto &f = field[ny][nx];
                    for (auto c : f)
                    {
                        if ( hit(c,p) )
                            ok = false;
                    }
                }
                if (ok)
                    field[by][bx].push_back(p);
            }
            ll cnt = 0;
            REP(i,maxBX+1)
            REP(j,maxBY+1)
            {
                auto &f = field[j][i];
                cnt += f.size();
            }
            cout<<cnt<<endl;
    }
};