2015 HIAST Collegiate Programming Contest J

题意：给2个凸多边形，求相交面积
思路：不会，套板子就是了
AC代码：
#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;

///2015 HIAST Collegiate Programming Contest

///JJJJ

#define maxn 510
const double eps=1E-8;
int sig(double d){
    return(d>eps)-(d<-eps);
}
struct Point{
    double x,y; Point(){}
    Point(double x,double y):x(x),y(y){}
    bool operator==(const Point&p)const{
        return sig(x-p.x)==0&&sig(y-p.y)==0;
    }
};
double cross(Point o,Point a,Point b){
    return(a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
}
double area(Point* ps,int n){
    ps[n]=ps[0];
    double res=0;
    for(int i=0;i<n;i++){
        res+=ps[i].x*ps[i+1].y-ps[i].y*ps[i+1].x;
    }
    return res/2.0;
}
int lineCross(Point a,Point b,Point c,Point d,Point&p){
    double s1,s2;
    s1=cross(a,b,c);
    s2=cross(a,b,d);
    if(sig(s1)==0&&sig(s2)==0) return 2;
    if(sig(s2-s1)==0) return 0;
    p.x=(c.x*s2-d.x*s1)/(s2-s1);
    p.y=(c.y*s2-d.y*s1)/(s2-s1);
    return 1;
}
//多边形切割
//用直线ab切割多边形p，切割后的在向量(a,b)的左侧，并原地保存切割结果
//如果退化为一个点，也会返回去,此时n为1
void polygon_cut(Point*p,int&n,Point a,Point b){
    static Point pp[maxn];
    int m=0;p[n]=p[0];
    for(int i=0;i<n;i++){
        if(sig(cross(a,b,p[i]))>0) pp[m++]=p[i];
        if(sig(cross(a,b,p[i]))!=sig(cross(a,b,p[i+1])))
            lineCross(a,b,p[i],p[i+1],pp[m++]);
    }
    n=0;
    for(int i=0;i<m;i++)
        if(!i||!(pp[i]==pp[i-1]))
            p[n++]=pp[i];
    while(n>1&&p[n-1]==p[0])n--;
}
//---------------华丽的分隔线-----------------//
//返回三角形oab和三角形ocd的有向交面积,o是原点//
double intersectArea(Point a,Point b,Point c,Point d){
    Point o(0,0);
    int s1=sig(cross(o,a,b));
    int s2=sig(cross(o,c,d));
    if(s1==0||s2==0)return 0.0;//退化，面积为0
    if(s1==-1) swap(a,b);
    if(s2==-1) swap(c,d);
    Point p[10]={o,a,b};
    int n=3;
    polygon_cut(p,n,o,c);
    polygon_cut(p,n,c,d);
    polygon_cut(p,n,d,o);
    double res=fabs(area(p,n));
    if(s1*s2==-1) res=-res;return res;
}
//求两多边形的交面积
double intersectArea(Point*ps1,int n1,Point*ps2,int n2){
    if(area(ps1,n1)<0) reverse(ps1,ps1+n1);
    if(area(ps2,n2)<0) reverse(ps2,ps2+n2);
    ps1[n1]=ps1[0];
    ps2[n2]=ps2[0];
    double res=0;
    for(int i=0;i<n1;i++){
        for(int j=0;j<n2;j++){
            res+=intersectArea(ps1[i],ps1[i+1],ps2[j],ps2[j+1]);
        }
    }
    return res;//assumeresispositive!
}
//hdu-3060求两个任意简单多边形的并面积
Point ps1[maxn],ps2[maxn];
int n1,n2;
int main(){
    int t;
    cin>>t;
    while(t--){
        scanf("%d%d",&n1,&n2);
        for(int i=0;i<n1;i++)
            scanf("%lf%lf",&ps1[i].x,&ps1[i].y);
        for(int i=0;i<n2;i++)
            scanf("%lf%lf",&ps2[i].x,&ps2[i].y);
        double ans=intersectArea(ps1,n1,ps2,n2);
        //ans=fabs(area(ps1,n1))+fabs(area(ps2,n2))-ans;//容斥
        printf("%.4f
",ans);
    }
    return 0;
}