【CQOI2006】凸多边形 计算几何

题目描述

　　逆时针给出n个凸多边形的顶点坐标，求它们交的面积。例如n=2时，两个凸多边形如下图：

　　则相交部分的面积为5.233。

数据范围

2<=n<=10，3<=mi<=50，每维坐标为[-1000,1000]内的整数

样例输入

2
6
-2 0
-1 -2
1 -2
2 0
1 2
-1 2
4
0 -3
1 -1
2 2
-1 0

样例输出

5.233

解题思路

半平面交，用每一条边去割原来的图形

代码

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include <ctime>#define Maxn 55using namespace std;struct Point{    double x,y;}a[Maxn],b[Maxn],c[Maxn];int m,n,nn;void Init(){    scanf("%d%d",&m,&n);    for(int i=1;i<=n;i++)        scanf("%lf%lf",&a[i].x,&a[i].y);}double Cross(Point a,Point b,Point c){    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}void AddCross(Point a,Point b,Point C,Point d){    double c1=Cross(a,d,b),c2=Cross(a,C,b);    c[++nn]=(Point){(c1*C.x-c2*d.x)/(c1-c2),(c1*C.y-c2*d.y)/(c1-c2)};}void Cut(Point A,Point B){    nn=0;    a[n+1]=a[1];    for(int i=1;i<=n;i++){        if(Cross(a[i],B,A)<=0){            c[++nn]=a[i];            if(Cross(a[i+1],B,A)>0)AddCross(A,B,a[i+1],a[i]);        }else if(Cross(a[i+1],B,A)<0)AddCross(A,B,a[i+1],a[i]);    }    for(int i=1;i<=nn;i++)a[i]=c[i];    n=nn;}int main(){    Init();    while(m---1){//233        int t;        scanf("%d",&t);        for(int i=1;i<=t;i++)            scanf("%lf%lf",&b[i].x,&b[i].y);        b[t+1]=b[1];        for(int i=1;i<=t;i++)            Cut(b[i],b[i+1]);    }    double Ans=0;    a[n+1]=a[1];    for(int i=1;i<=n;i++)        Ans+=a[i].x*a[i+1].y-a[i].y*a[i+1].x;    printf("%.3lf",fabs(Ans)/2);}