• codeforces1468F Full Turn


    题意

    给定(n)个点(p_i=(x_i, y_i)),第(i)个点初始时朝向((tx_i, ty_i))。每个点的朝向以相同的角速度转动,转满一圈停止。

    (A)和点(B)对视,即(B)在点(A)朝向的方向上,且(A)在点(B)朝向的方向上。两点中间有其他点不会影响对视。

    问满足某一时刻两点对视的点对数。

    解法

    性质1:若某一时刻,(A)(B)对视,则此时两点朝向的夹角为(180)度。

    性质2:因为角速度相同,所以任意时刻,两点朝向的夹角等于初始时的夹角。

    根据这两点性质,可以推出满足某一时刻两点对视的点对数即为初始时朝向夹角为(180)度的点对数。

    然后因为可能会有精度问题,所以并不能直接用三角函数来算夹角。

    令第(p_i)的方向向量为(vec d_i = (tx_i - x_i, ty_i - y_i) = (dx_i, dy_i))。若将(vec d_i)标准化,则(p_i)(p_j)初始朝向夹角为(180)度当且仅当(vec d_i + vec d_j = vec 0)

    标准化的方式就是,若(vec d_i)水平或者竖直,就令(vec d_i = frac{vec d_i}{|vec d_i|});否则,就让(vec d_i = (frac{dx_i}{g}, frac{dy_i}{g})),其中(g = GCD(dx_i, dy_i))

    然后可以用一个map乱搞完事了。

    AC代码

    #include <bits/stdc++.h>
    using namespace std;
     
    using ll = int64_t;
    using ull = uint64_t;
    using uint = uint32_t;
    using VI = vector<int>;
    using VL = vector<ll>;
    using VVI = vector<vector<int>>;
    using VVL = vector<vector<ll>>;
    using PII = pair<int,int>;
    using PLL = pair<ll, ll>;
     
    #define REP(i, _, __) for (int i = (_); i < (__); ++i)
    #define PER(i, _, __) for (int i = (_-1); i >= (__); --i)
    #define FOR(i, _, __) for (int i = (_); i <= (__); ++i)
    #define ROF(i, _, __) for (int i = (_); i >= (__); --i)
    #define FC(v, V) for (const auto& v: V)
    #define FE(v, V) for (auto& v: V)
    
    #define EB emplace_back
    #define PB push_back
    #define MP make_pair
    #define FI first
    #define SE second
    #define SZ(x) (int((x).size()))
    #define ALL(x) (x).begin(),(x).end()
    #define LLA(x) (x).rbegin(),(x).rend()
    
    const double PI = acos(-1.0);
       
    namespace Backlight {
        const int __BUFFER_SIZE__ = 1 << 20;
        bool NEOF = 1;
        int __top;
        char __buf[__BUFFER_SIZE__], *__p1 = __buf, *__p2 = __buf, __stk[996];
    
        template<typename T>
        T MIN(T a, T b) { return min(a, b); }
    
        template<typename First, typename... Rest>
        First MIN(First f, Rest... r) { return min(f, MIN(r...)); }
    
        template<typename T>
        T MAX(T a, T b) { return max(a, b); }
    
        template<typename First, typename... Rest>
        First MAX(First f, Rest... r) { return max(f, MAX(r...)); }
    
        template<typename T>
        void updMin(T& a, T b) { if (a > b) a = b; }
    
        template<typename T>
        void updMax(T& a, T b) { if (a < b) a = b; }
    
        inline char nc() {
            return __p1 == __p2 && NEOF && (__p2 = (__p1 = __buf) + fread(__buf, 1, __BUFFER_SIZE__, stdin), __p1 == __p2) ? (NEOF = 0, EOF) : *__p1++;
        }
       
        template<typename T>
        inline bool read(T &x) {
            char c = nc();
            bool f = 0; x = 0;
            while (!isdigit(c)) c == '-' && (f = 1), c = nc();
            while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = nc();
            if (f) x = -x;
            return NEOF;
        }
    
        inline bool need(char c) { return (c != '
    ') && (c != ' '); }
    
        inline bool read(char& a) {
            while ((a = nc()) && need(a) && NEOF) ;
            return NEOF;
        }
    
        inline bool read(char *a) {
            while ((*a = nc()) && need(*a) && NEOF) ++a; 
            *a = '';
            return NEOF;
        }
    
        inline bool read(double &x) {
            bool f = 0; char c = nc(); x = 0;
            while (!isdigit(c))  { f |= (c == '-'); c = nc(); }
            while (isdigit(c)) { x = x * 10.0 + (c ^ 48); c = nc(); }
            if (c == '.') {
                double temp = 1; c = nc();
                while (isdigit(c)) { temp = temp / 10.0; x = x + temp * (c ^ 48); c = nc(); }
            }
            if (f) x = -x;
            return NEOF;
        }
    
        template<typename First, typename... Rest>
        inline bool read(First &f, Rest &... r) {
            read(f);
            return read(r...);
        }
    
        template<typename T>
        inline void print(T x) {
            if (x < 0) putchar('-'), x = -x;
            if (x == 0) { putchar('0'); return; }
            __top = 0;
            while(x) {
                __stk[++__top] = x % 10 + '0';
                x /= 10;
            }
            while(__top) {
                putchar(__stk[__top]);
                --__top;
            }
        }
    
        template<typename First, typename... Rest>
        inline void print(First f, Rest... r) {
            print(f); putchar(' ');
            print(r...);
        }
    
        template<typename T>
        inline void println(T x) {
            print(x); 
            putchar('
    ');
        }
    
        template<typename First, typename... Rest>
        inline void println(First f, Rest... r) {
            print(f); putchar(' ');
            println(r...);
        }
    
        template<typename T>
        inline void _dbg(const char *format, T value) { cerr << format << '=' << value << endl; }
       
        template<typename First, typename... Rest>
        inline void _dbg(const char *format, First f, Rest... r) {
            while(*format != ',') cerr << *format++;
            cerr << '=' << f << ", ";
            _dbg(format + 1, r...);
        }
          
        template<typename T>
        ostream &operator<<(ostream& os, vector<T> V) {
            os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
        }
       
        template<typename T>
        ostream &operator<<(ostream& os, set<T> V) {
            os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
        }
    
        template<typename T>
        ostream &operator<<(ostream& os, multiset<T> V) {
            os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
        }
     
        template<typename T1, typename T2>
        ostream &operator<<(ostream& os, map<T1, T2> V) {
            os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
        }
      
        template<typename L, typename R>
        ostream &operator<<(ostream &os, pair<L, R> P) {
            return os << "(" << P.first << "," << P.second << ")";
        }
    
        #ifdef BACKLIGHT
        #define debug(...) cerr << "33[31m" << "[" << __LINE__ << "] : "; _dbg(#__VA_ARGS__, __VA_ARGS__); cerr << "33[0m";
        // #define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__); 
        #else
        #define debug(...)
        #endif
    }
    
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    int rnd(int l, int r) { return l + rng() % (r - l + 1); }
    
    using namespace Backlight;
    const int N = 3e5 + 5;
    const int M = 3e6 + 5;
    const int K = 1e7 + 5;
    const int MOD = 1e9 + 7;              // 998244353 1e9 + 7
    const int INF = 0x3f3f3f3f;             // 1e9 + 7 0x3f3f3f3f
    const ll LLINF = 0x3f3f3f3f3f3f3f3f;    // 1e18 + 9 0x3f3f3f3f3f3f3f3f
    const double eps = 1e-8;
    
    struct point {
        int x, y;
        point operator + (const point& p) const {
            return (point){x + p.x, y + p.y};
        }
        bool operator < (const point& p) const {
            return x < p.x || (x == p.x && y < p.y);
        }
        void input() {
            int sx, sy, tx, ty;
            read(sx, sy);
            read(tx, ty);
            x = tx - sx; y = ty - sy;
            if (x == 0 && y == 0);
            else if (x == 0) y = y / abs(y);
            else if (y == 0) x = x / abs(x);
            else {
                int g = abs(__gcd(x, y));
                x /= g;
                y /= g;
            }
    
            debug(x, y);
        }
        point inv() {
            return (point){-x, -y};
        }
    };
    // true if alphta - eta = 180 degree
    
    point p[N];
    void solve(int Case) { // printf("Case #%d: ", Case);
        int n;
        read(n);
    
        FOR(i, 1, n) p[i].input();
        
        ll ans = 0;
    
        map<int, int> cx, cy;
        map<point, int> cp;
        FOR(i, 1, n) {
            ans += cp[p[i].inv()];
    
            ++cp[p[i]];
    
            debug(cx, cy);
        }
    
        println(ans);
    }
    
    
    int main() {
    #ifdef BACKLIGHT
        freopen("in.txt", "r", stdin);
        // freopen("out.txt", "w", stdout);
        auto begin = std::chrono::steady_clock::now();
    #endif
    
        // ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
        int T = 1;
        read(T);
        for (int _ = 1; _ <= T; _++) solve(_);
    
    #ifdef BACKLIGHT
        auto end = std::chrono::steady_clock::now();
        auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(end - begin);
        cerr << "33[32mTime Elasped: " << duration.count() << " ms33[0m" << endl;
    #endif
        return 0;
    }
    
  • 相关阅读:
    Android开发开源一款结合databinding写的用于RecyclerView的简单高效MultiTypeAdapter
    Android开发databinding和RecyclerView.ViewHolder的完美结合
    Android开发华为手机不弹出Toast,报HwRTBlurUtils: check blur style for HwToast-Toast...的原因
    android开发Toolbar标题居中显示的解决方法
    记录使用xshell通过ssh方式连接Linux机器的步骤
    同一局域网内手机访问电脑本地localhost网页的方法
    Gradle里面的依赖implementation和api的真正理解
    Android开发使用kotlin编写的泛型模式的MVP框架
    nyoj-3-多边形重心问题(求多边形面积和中心)
    nyoj-1132-promise me a medal(求线段交点)
  • 原文地址:https://www.cnblogs.com/zengzk/p/14192112.html
Copyright © 2020-2023  润新知