[博弈论][凸包][复杂度分析] SRM 597 Div1 Medium ConvexPolygonGame

最新推荐文章于 2022-11-28 11:14:07 发布

Vectorxj

最新推荐文章于 2022-11-28 11:14:07 发布

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分类专栏：复杂度分析计算几何博弈论

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本文介绍了一种判断凸多边形游戏中胜负的方法。通过分析凸多边形的特性，利用O(n+X)的时间复杂度来确定游戏的胜负方。首先检查是否存在三点不共线的情况，然后通过扫描特定直线与凸多边形的交集进一步判断。

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$Solution$

设 $W_i$ 为第 $i$ 次操作后可以得到的凸多边形的集合。
显然有 $W_i\subseteq W_j,i\gt j$ 。
所以如果最开始不能操作则先手必败，否则先手必胜。
问题就在于判断凸多边形中是否存在三点不共线。
暴力判断是 $O(X^2)$ 的。
转化成任意三点共线，这样的点数不超过 $max(max_X - min_X,max_Y - min_Y)$ 。扫描直线 $x=X$ 与凸多边形的交的部分，若个数超过上述限制则先手必胜。
否则判断直线 $x=X$ 与凸多边形的交的整点中是否存在三点不共线。
复杂度 $O(n+X)$ 。

// BEGIN CUT HERE

// END CUT HERE
#line 5 "ConvexPolygonGame.cpp"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> Pairs;
const int N = 101010;
const double eps = 1e-6;

Pairs p[N];
int n, lim, p1, p2, st, mn, B;
int downY[N << 1];
vector<Pairs> pp;

inline Pairs operator -(Pairs a, Pairs b) {
    return Pairs(a.first - b.first, a.second - b.second);
}

class ConvexPolygonGame {

public:
    inline ll Det(Pairs a, Pairs b) {
        return (ll)a.first * b.second - (ll)a.second * b.first;
    }
    string winner(vector <int> X, vector <int> Y) {
        n = X.size(); mn = lim = 2000001; B = 100000;
        for (int i = 0; i < n; i++) {
            p[i] = Pairs(X[i], Y[i]);
            if (X[i] < mn || (X[i] == mn && Y[i] < Y[st])) {
                mn = X[i]; st = i;
            }
        }
        p[n] = p[n - 1];
        for (int d = 0; d < n; d++) {
            Pairs bg = p[(st + d) % n], ed = p[(st + d + 1) % n];
            if (bg > ed) {
                swap(bg, ed);
                if (bg.first == ed.first) {
                    for (int y = bg.second + 1; y < ed.second; y++)
                        pp.push_back(Pairs(bg.first, y));
                } else {
                    for (int x = bg.first; x <= ed.first; x++) {
                        double k = (double)(bg.second - ed.second) / (bg.first - ed.first);
                        double b = ed.second - k * ed.first;
                        int upY = floor(k * x + b + eps);
                        for (int y = downY[x + B]; y <= upY; y++)
                            if (Pairs(x, y) != bg && Pairs(x, y) != ed)
                                pp.push_back(Pairs(x, y));
                        if (pp.size() > lim) return "Masha";
                    }
                }
            } else {
                if (bg.first == ed.first) {
                    downY[bg.first + B] = bg.second + 1;
                } else {
                    for (int x = bg.first; x <= ed.first; x++) {
                        double k = (double)(bg.second - ed.second) / (bg.first - ed.first);
                        double b = ed.second - k * ed.first;
                        int y = ceil(k * x + b - eps);
                        if (Pairs(x, y) == bg || Pairs(x, y) == ed) ++y;
                        downY[x + B] = y;
                        if (pp.size() > lim) return "Masha";
                    }
                }
            }
            if (pp.size() > lim) return "Masha";
        }
        for (int i = 2; i < pp.size(); i++)
            if (Det(pp[0] - pp[1], pp[1] - pp[i])) return "Masha";
        return "Petya";
    }

// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); }
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
    void verify_case(int Case, const string &Expected, const string &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
    void test_case_0() { int Arr0[] = {0, 1, 0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0, 0, 1}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arg2 = "Petya"; verify_case(0, Arg2, winner(Arg0, Arg1)); }
    void test_case_1() { int Arr0[] = {0, 4, 2}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0, 0, 2}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arg2 = "Masha"; verify_case(1, Arg2, winner(Arg0, Arg1)); }
    void test_case_2() { int Arr0[] = {0, 100, 100, 0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0, 0, 100, 100}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arg2 = "Masha"; verify_case(2, Arg2, winner(Arg0, Arg1)); }
    void test_case_3() { int Arr0[] = {0, 50, 100, 50}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0, -1, 0, 1}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arg2 = "Petya"; verify_case(3, Arg2, winner(Arg0, Arg1)); }
    void test_case_4() { int Arr0[] = {-100000, 100000, 100000, -100000}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {-1, -1, 1, 1}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arg2 = "Masha"; verify_case(4, Arg2, winner(Arg0, Arg1)); }
    void test_case_5() { int Arr0[] = {-100000, 100000, 100000, -100000}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {-100000, -100000, 100000, 100000}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arg2 = "Masha"; verify_case(5, Arg2, winner(Arg0, Arg1)); }

// END CUT HERE


};

// BEGIN CUT HERE
int main(void) {
    ConvexPolygonGame ___test;
    ___test.run_test(5);
    system("pause");
}
// END CUT HERE