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CSU 1812 三角形和矩形

时间:2016-09-03 23:55:55      阅读:517      评论:0      收藏:0      [点我收藏+]

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湖南省第十二届大学生计算机程序设计竞赛$J$题

计算几何。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c))
    {
        x = x * 10 + c - 0;
        c = getchar();
    }
}

const int maxn=555;
const int maxisn=10;

int dcmp(double x)
{
    if(x>eps) return 1;
    return x<-eps ? -1 : 0;
}
inline double Sqr(double x)
{
    return x*x;
}
struct Point
{
    double x,y;
    Point()
    {
        x=y=0;
    }
    Point(double x,double y):x(x),y(y) {};
    friend Point operator + (const Point &a,const Point &b)
    {
        return Point(a.x+b.x,a.y+b.y);
    }
    friend Point operator - (const Point &a,const Point &b)
    {
        return Point(a.x-b.x,a.y-b.y);
    }
    friend bool operator == (const Point &a,const Point &b)
    {
        return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
    }
    friend Point operator * (const Point &a,const double &b)
    {
        return Point(a.x*b,a.y*b);
    }
    friend Point operator * (const double &a,const Point &b)
    {
        return Point(a*b.x,a*b.y);
    }
    friend Point operator / (const Point &a,const double &b)
    {
        return Point(a.x/b,a.y/b);
    }
    friend bool operator < (const Point &a, const Point &b)
    {
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    }
    inline double dot(const Point &b)const
    {
        return x*b.x+y*b.y;
    }
    inline double cross(const Point &b,const Point &c)const
    {
        return (b.x-x)*(c.y-y)-(c.x-x)*(b.y-y);
    }

};

Point LineCross(const Point &a,const Point &b,const Point &c,const Point &d)
{
    double u=a.cross(b,c),v=b.cross(a,d);
    return Point((c.x*v+d.x*u)/(u+v),(c.y*v+d.y*u)/(u+v));
}
double PolygonArea(Point p[],int n)
{
    if(n<3) return 0.0;
    double s=p[0].y*(p[n-1].x-p[1].x);
    p[n]=p[0];
    for(int i=1; i<n; i++)
    {
        s+=p[i].y*(p[i-1].x-p[i+1].x);
    }
    return fabs(s*0.5);
}
double CPIA(Point a[],Point b[],int na,int nb)
{
    Point p[maxisn],temp[maxisn];
    int i,j,tn,sflag,eflag;
    a[na]=a[0],b[nb]=b[0];
    memcpy(p,b,sizeof(Point)*(nb+1));
    for(i=0; i<na&&nb>2; ++i)
    {
        sflag=dcmp(a[i].cross(a[i+1],p[0]));
        for(j=tn=0; j<nb; ++j,sflag=eflag)
        {
            if(sflag>=0) temp[tn++]=p[j];
            eflag=dcmp(a[i].cross(a[i+1],p[j+1]));
            if((sflag^eflag)==-2)
                temp[tn++]=LineCross(a[i],a[i+1],p[j],p[j+1]);
        }
        memcpy(p,temp,sizeof(Point)*tn);
        nb=tn,p[nb]=p[0];
    }
    if(nb<3) return 0.0;
    return PolygonArea(p,nb);
}
double SPIA(Point a[],Point b[],int na,int nb)
{
    int i,j;
    Point t1[4],t2[4];
    double res=0.0,if_clock_t1,if_clock_t2;
    a[na]=t1[0]=a[0];
    b[nb]=t2[0]=b[0];
    for(i=2; i<na; i++)
    {
        t1[1]=a[i-1],t1[2]=a[i];
        if_clock_t1=dcmp(t1[0].cross(t1[1],t1[2]));
        if(if_clock_t1<0) swap(t1[1],t1[2]);
        for(j=2; j<nb; j++)
        {
            t2[1]=b[j-1],t2[2]=b[j];
            if_clock_t2=dcmp(t2[0].cross(t2[1],t2[2]));
            if(if_clock_t2<0) swap(t2[1],t2[2]);
            res+=CPIA(t1,t2,3,3)*if_clock_t1*if_clock_t2;
        }
    }
    return res;
}

Point a[222],b[222];
Point aa[222],bb[222];

int main()
{
    int n1,n2;
    double x1,y1,x2,y2,x3,y3,x4,y4;
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4))
    {
        n1=3,n2=4;
        a[0].x=x1; a[0].y=y1;
        a[1].x=x1; a[1].y=y2;
        a[2].x=x2; a[2].y=y1;

        b[0].x=x3; b[0].y=y3;
        b[1].x=x3; b[1].y=y4;
        b[2].x=x4; b[2].y=y4;
        b[3].x=x4; b[3].y=y3;

        printf("%.6lf\n",abs(SPIA(a,b,n1,n2)));
    }
    return 0;
}

 

CSU 1812 三角形和矩形

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原文地址:http://www.cnblogs.com/zufezzt/p/5838163.html

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