csu 1812: 三角形和矩形 凸包

Posted 2020-08-09 youmi

传送门：csu 1812: 三角形和矩形

思路：首先，求出三角形的在矩形区域的顶点，矩形在三角形区域的顶点。然后求出所有的交点。这些点构成一个凸包，求凸包面积就OK了。 

/**************************************************************
    Problem:
    User: youmi
    Language: C++
    Result: Accepted
    Time:
    Memory:
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%I64d",&a)
#define pt(a) printf("%d\n",a)
#define ptlld(a) printf("%I64d\n",a)
#define rep(i,from,to) for(int i=from;i<=to;i++)
#define irep(i,to,from) for(int i=to;i>=from;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define eps 1e-6
#define oo 0x3fffffff
#define TEST cout<<"*************************"<<endl
const double pi=4*atan(1.0);

using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
    char c; int flag = 1;
    for (c = getchar(); !(c >= ‘0‘ && c <= ‘9‘ || c == ‘-‘); c = getchar()); if (c == ‘-‘) flag = -1, n = 0; else n = c - ‘0‘;
    for (c = getchar(); c >= ‘0‘ && c <= ‘9‘; c = getchar()) n = n * 10 + c - ‘0‘; n *= flag;
}
ll Pow(ll base, ll n, ll mo)
{
    if (n == 0) return 1;
    if (n == 1) return base % mo;
    ll tmp = Pow(base, n >> 1, mo);
    tmp = (ll)tmp * tmp % mo;
    if (n & 1) tmp = (ll)tmp * base % mo;
    return tmp;
}
//***************************

int n;
const int maxn=100000+10;
const ll mod=1000000007;
double xx[10],yy[10];
int sgn(double x)
{
    if(fabs(x)<eps)
        return 0;
    if(x<0)
        return -1;
    else
        return 1;
}
struct point
{
    double x,y;
    point(){};
    point(double _x,double _y)
    {
        x=_x,y=_y;
    }
    point operator-(const point &_b)const
    {
        return point(x-_b.x,y-_b.y);
    }
    double operator *(const point &_b)const
    {
        return x*_b.x+y*_b.y;
    }
    double operator^(const point &_b)const
    {
        return x*_b.y-_b.x*y;
    }
    bool operator==(const point &_b)const
    {
        return sgn(x-_b.x)==0&&sgn(y-_b.y)==0;
    }
};
point tri[10],rec[10];
double dist(point a,point b)
{
    return sqrt((a-b)*(a-b));
}
struct line
{
    point s, e;
    line() {}
    line(point _s, point _e)
    {
        s = _s;
        e = _e;
    }
    pair<int, point> operator &(const line &b)const
    {
        point res = s;
        if(sgn((s - e) ^ (b.s - b.e)) == 0) {
            if(sgn((s - b.e) ^ (b.s - b.e)) == 0)
                return make_pair(0, res); //重合
            else return make_pair(1, res); //平行
        }
        long double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e));
        res.x += (e.x - s.x) * t;
        res.y += (e.y - s.y) * t;
        return make_pair(2, res);
    }
};
point lst[maxn];
int stc[maxn],top;
bool _cmp(point p1,point p2)
{
    double temp=(p1-lst[0])^(p2-lst[0]);
    if(sgn(temp)>0)
        return true;
    else if(sgn(temp)==0&&sgn(dist(p1,lst[0])-dist(p2,lst[0]))<=0)
        return true;
    else
        return false;
}
bool on_line(point p,line uu)
{
    return (sgn(p.x-uu.s.x)*sgn(p.x-uu.e.x))<=0&&(sgn(p.y-uu.s.y)*sgn(p.y-uu.e.y)<=0);
}
void graham()
{
    if(n==0)
    {
        top=0;
        return;
    }
    point p0=lst[0];
    int k=0;
    for(int i=1;i<n;i++)
    {
        if((p0.y>lst[i].y)||(p0.y==lst[i].y&&p0.x>lst[i].x))
        {
            p0=lst[i];
            k=i;
        }
    }
    swap(lst[k],lst[0]);
    sort(lst+1,lst+n,_cmp);
    if(n==1)
    {
        top=1;
        stc[0]=0;
        return ;
    }
    top=2;
    stc[0]=0;
    stc[1]=1;
    if(n==2)
        return ;
    for(int i=2;i<n;i++)
    {
        while(top>1&&sgn((lst[stc[top-1]]-lst[stc[top-2]])^(lst[i]-lst[stc[top-2]]))<=0)
            top--;
        stc[top++]=i;
    }
}
bool in_tri(point p)
{
    double s=fabs((tri[0]-tri[1])^(tri[0]-tri[2]));
    double s1=fabs((p-tri[0])^(p-tri[1]));
    double s2=fabs((p-tri[1])^(p-tri[2]));
    double s3=fabs((p-tri[2])^(p-tri[0]));
    return sgn(s1+s2+s3-s)==0;
}
bool in_rec(point p)
{
    return sgn(p.x-xx[3])>=0&&sgn(p.x-xx[4])<=0&&sgn(p.y-yy[3])>=0&&sgn(p.y-yy[4])<=0;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",xx+1,yy+1,xx+2,yy+2,xx+3,yy+3,xx+4,yy+4))
    {
        tri[0]=point(xx[1],yy[1]),tri[1]=point(xx[1],yy[2]),tri[2]=point(xx[2],yy[1]);
        rec[0]=point(xx[3],yy[3]),rec[1]=point(xx[3],yy[4]),rec[2]=point(xx[4],yy[4]),rec[3]=point(xx[4],yy[3]);
        n=0;
        rep(i,0,3)
            if(in_tri(rec[i]))
                lst[n++]=rec[i];
        rep(i,0,2)
            if(in_rec(tri[i]))
                lst[n++]=tri[i];
        rep(i,0,3)
        {
            line gg=line(rec[i],rec[(i+1)%4]);
            rep(j,0,2)
            {
                line hh=line(tri[j],tri[(j+1)%3]);
                pair<int, point> res=hh&gg;
                if(res.first==2)
                {
                    point &uu=res.second;
                    if(on_line(uu,gg)&&on_line(uu,hh))
                        lst[n++]=uu;
                }
            }
        }
        sort(lst,lst+n,_cmp);
        n=unique(lst,lst+n)-lst;
        graham();
        double ans=0;
        if(top>=3)
        {
            rep(i,0,top-1)
            {
                ans+=lst[stc[i]]^lst[stc[(i+1)%top]];
            }
        }
        ans/=2;
        printf("%.7f\n",ans);
    }
}