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  • Boundary

    多校赛第二场B题

    #include <bits/stdc++.h>
    #define ll              long long
    #define lll             unsigned long long
    #define pii             pair<int, int>
    #define rep(i,a,b)      for(ll  i=a;i<=b;i++)
    #define dec(i,a,b)      for(ll  i=a;i>=b;i--)
    #define forn(i, n)      for(ll i = 0; i < int(n); i++)
    using namespace std;
    int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = 3.14159265358979323846;
    const double eps = 1e-6;
    const int mod = 998244353;
    const int N = 2e3 + 5;
    //if(x<0 || x>=r || y<0 || y>=c)
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    ll lcm(ll m, ll n)
    {
        return m * n / gcd(m, n);
    }
    bool prime(int x) {
        if (x < 2) return false;
        for (int i = 2; i * i <= x; ++i) {
            if (x % i == 0) return false;
        }
        return true;
    }
    inline int qpow(int x, ll n) {
        int r = 1;
        while (n > 0) {
            if (n & 1) r = 1ll * r * x % mod;
            n >>= 1; x = 1ll * x * x % mod;
        }
        return r;
    }
    inline int add(int x, int y) {
        return ((x % mod) + (y % mod)) % mod;
    }
    inline int sub(int x, int y) {
        x -= y;
        return x < 0 ? x += mod : x;
    }
    inline int mul(int x, int y) {
        return (1ll * (x % mod) * (y % mod)) % mod;
    }
    inline int inv(int x) {
        return qpow(x, mod - 2);
    }
    inline int divd(int x, int y) {
        return (1ll * (x % mod) * inv(y)) % mod;
    }
    int n, ans = 1, X[N], Y[N];
    
    struct node
    {
        double x,y;
        node(): node(0,0){}
        node(double x,double y):x(x),y(y){}
        bool operator < (const node& r)
        {
            return (double)x*r.y<(double)y*r.x;
        }
        bool operator == (const node& r)
        {
            return (double)x*r.y==(double)y*r.x;
        }
    };
    
    int Cross(int lhs, int rhs)
    {
        return X[lhs] * Y[rhs] - X[rhs] * Y[lhs];
    }
    
    int Dot(int lhs, int rhs)
    {
        return X[lhs] * X[rhs] + Y[lhs] * Y[rhs];
    }
    
    double Dis2(int lhs, int rhs)
    {
        double dx = X[lhs] - X[rhs], dy = Y[lhs] - Y[rhs];
        return dx * dx + dy * dy;
    }
    
    int Sgn(int x)
    {
        if (x > 0) return 1;
        if (x < 0) return -1;
        return 0;
    }
    
    node ar[N];
    
    node cos2(ll x, ll y)
    {
        ll a2=Dis2(0,x),b2=Dis2(x,y),c2=Dis2(0,y);
        int sgn=Sgn(b2+c2-a2);
        return node((double)1*sgn*(b2 + c2 - a2) * (b2 + c2 - a2), (double)4 * b2 * c2);
    }
    
    int main()
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i ++)
            scanf("%d%d", X + i, Y + i);
        for (int i = 1; i <= n; i ++)
        {
            int cnt = 0;
            for (int j = 1; j <= n; j ++)
                if (Cross(i, j) > 0)
                    ar[++cnt]=cos2(i,j);
            sort(ar+1,ar+cnt+1);
            for(int l=1,r;l<=cnt;l=r)
            {
                for(r=l;ar[l]==ar[r] && r<=cnt;r++);
                ans=max(ans,r-l+1);
            }
        }
        printf("%d
    ", ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dealer/p/13320729.html
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