HDU 4082 Hou Yi's secret --枚举

题意： 给n个点，问最多有多少个相似三角形（三个角对应相等）。

解法： O(n^3)枚举点，形成三角形，然后记录三个角，最后按三个角度依次排个序，算一下最多有多少个连续相等的三元组就可以了。

注意：在同一个坐标的两点只算一次，所以要判一下。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#define eps 1e-8

using namespace std;

#define N 100017



struct Point{

    double x,y;

    Point(double x=0, double y=0):x(x),y(y) {}

    void input() { scanf("%lf%lf",&x,&y); }

};

typedef Point Vector;

int dcmp(double x) {

    if(x < -eps) return -1;

    if(x > eps) return 1;

    return 0;

}

template <class T> T sqr(T x) { return x * x;}

Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }

Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }

Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }

Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }

bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }

bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }

bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }

double Length(Vector A) { return sqrt(Dot(A, A)); }

double Angle(Vector A, Vector B) { return (Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }



//data segment

struct Tri{

    double A[3];

    Tri(double x,double y,double z)

    { A[0] = x, A[1] = y, A[2] = z; }

    Tri(){}

    bool operator <(const Tri& a)const

    {

        if(dcmp(A[0]-a.A[0]) == 0)

        {

            if(dcmp(A[1]-a.A[1])==0) return dcmp(A[2]-a.A[2])<0;

            return dcmp(A[1]-a.A[1])<0;

        }

        return dcmp(A[0]-a.A[0])<0;

    }

}t[3005];

bool operator == (const Tri& a,const Tri& b) { return dcmp(a.A[0]-b.A[0]) == 0 && dcmp(a.A[1]-b.A[1]) == 0 && dcmp(a.A[2]-b.A[2]) == 0; }

Point p[25];

int tot,n;

//data ends

int mp[300][300];

int main()

{

    int i,j,k;

    int a[5];

    while(scanf("%d",&n)!=EOF && n)

    {

        memset(mp,0,sizeof mp);

        tot = 0;

        int cntt=1;

        for(i=1;i<=n;i++)

        {

            int a,b;scanf("%d%d",&a,&b);

            if(mp[a+100][b+100]==0)

                p[cntt].x=a,p[cntt++].y=b;

            mp[a+100][100+b]=1;

        }

        n=cntt-1;

        for(i=1;i<=n;i++)

        {

            for(j=i+1;j<=n;j++)

            {

                for(k=j+1;k<=n;k++)

                {

                    Point A = p[i], B = p[j], C = p[k];

                    if(A == B || A == C || B == C) continue;

                    if(dcmp(Cross(B-A,C-A)) == 0) continue;

                    double ang1 = Angle(B-A,C-A);

                    double ang2 = Angle(A-B,C-B);

                    double ang3 = Angle(A-C,B-C);

                    double A1 = min(ang1,min(ang2,ang3));

                    double A3 = max(ang1,max(ang2,ang3));

                    double A2 = ang1+ang2+ang3-A1-A3;

                    t[++tot] = Tri(A1,A2,A3);

                }

            }

        }

        sort(t+1,t+tot+1);

        int Maxi = (tot!=0), cnt = 1;

        for(i=2;i<=tot;i++)

        {

            if(t[i] == t[i-1])

                cnt++;

            else

                cnt = 1;

            Maxi = max(Maxi,cnt);

        }

        cout<<Maxi<<endl;

    }

    return 0;

}

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