hdu3060Area2(任意多边形相交面积)

链接

多边形的面积求解是通过选取一个点（通常为原点或者多边形的第一个点）和其它边组成的三角形的有向面积。

对于两个多边形的相交面积就可以通过把多边形分解为三角形，求出三角形的有向面积递加。三角形为凸多边形，因此可以直接用凸多边形相交求面积的模板。

凸多边形相交后的部分肯定还是凸多边形，所以只需要判断哪些点是相交部分上的点，最后求下面积。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 510
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y) {} //构造函数 方便代码编写
}p[N],q[N];
typedef point pointt;
pointt operator + (point a,point b)
{
    return point(a.x+b.x,a.y+b.y);
}
pointt operator - (point a,point b)
{
    return point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
bool operator == (const point &a,const point &b)
{
    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double cross(point a,point b)
{
    return a.x*b.y-a.y*b.x;
}
double Polyarea(point p[],int n)
{
    if(n<3) return 0;
    double area = 0;
    for(int i = 0 ; i < n ; i++)
    area+=cross(p[i],p[i+1]);
    return fabs(area)/2;
}
//double Polyarea(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);
//}
bool intersection1(point p1, point p2, point p3, point p4, point& p)      // 直线相交
{
    double a1, b1, c1, a2, b2, c2, d;
    a1 = p1.y - p2.y;
    b1 = p2.x - p1.x;
    c1 = p1.x*p2.y - p2.x*p1.y;
    a2 = p3.y - p4.y;
    b2 = p4.x - p3.x;
    c2 = p3.x*p4.y - p4.x*p3.y;
    d = a1*b2 - a2*b1;
    if (!dcmp(d))    return false;
    p.x = (-c1*b2 + c2*b1) / d;
    p.y = (-a1*c2 + a2*c1) / d;
    return true;
}

double convexpolygon(point p[],point q[],int n,int m)
{
    int i,j;
    point ch[20],cnt[20];
    p[n] = p[0],q[m] = q[0];
    memcpy(ch,q,sizeof(point)*(m+1));
    int f2,g = 0 ;
    for(i = 0 ;i < n; i++)
    {
        int f1 = dcmp(cross(p[i+1]-p[i],ch[0]-p[i]));
        g = 0 ;
        for(j = 0 ;j < m; j++,f1 = f2)
        {
            if(f1>=0) cnt[g++] = ch[j];
            f2 = dcmp(cross(p[i+1]-p[i],ch[j+1]-p[i]));
            if((f1^f2)==-2)
            {
                point pp;
                intersection1(p[i],p[i+1],ch[j],ch[j+1],pp);
                cnt[g++] = pp;
            }
        }
        cnt[g] = cnt[0];
        memcpy(ch,cnt,sizeof(point)*(g+1));
        m = g;
    }
    return Polyarea(ch,g);
}
double solve(point p[],point q[],int n,int m)
{
    int i,j;
    double area = 0;
    point tp[10],tq[10];
    tp[0] = p[0];tq[0] = q[0];
    for(i = 1 ;i < n-1; i++)
    {
        int k1 = dcmp(cross(p[i]-p[0],p[i+1]-p[0]));
        tp[1] = p[i],tp[2] = p[i+1];
        if(k1 < 0) swap(tp[1],tp[2]);

        for(j = 1 ;j < m-1 ;j++)
        {
            tq[1] = q[j],tq[2] = q[j+1];
            int k2 = dcmp(cross(q[j]-q[0],q[j+1]-q[0]));
            if(k2<0) swap(tq[1],tq[2]);

            area+=convexpolygon(tp,tq,3,3)*k1*k2;
        }
    }
    return Polyarea(p,n)+Polyarea(q,m)-area;
}
int main()
{
    int n,m,i;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i = 0; i < n ;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        for(i = 0; i < m; i++)
        {
            scanf("%lf%lf",&q[i].x,&q[i].y);
        }
        p[n] = p[0],q[m] = q[0];
        double ans = solve(p,q,n,m);
        printf("%.2f
",ans);
    }
    return 0;
}