UVA 11437 Triangle Fun

来源：互联网发布：每股收益的算法编辑：程序博客网时间：2024/06/07 06:22

题目链接：https://vjudge.net/problem/UVA-11437

题意：简单来说就是求S（PRQ）

自己做的时候一直认为三角形ABC和PRQ之间有什么等价关系，证明了很久还是证明不出来，后来发现完全可以硬怼出来，= =，直接计算出P,Q,R的坐标，根据叉乘就面积即可

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<queue>#include<deque>#include<set>#include<map>#include<cmath>#include<vector>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> PII;#define pi acos(-1.0)#define eps 1e-10#define pf printf#define sf scanf#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r#define epp tree[rt]#define _s second#define _f first#define all(x) (x).begin,(x).end#define mem(i,a) memset(i,a,sizeof i)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define mi ((l+r)>>1)#define sqr(x) ((x)*(x))struct Point{    double x,y;    Point(){}    Point(double x,double y):x(x),y(y){}}A,B,C,D,E,F,X,Y,Z;const int inf=0x3f3f3f3f;int t;Point getpoint(Point A,Point B,Point C,Point D)//计算交点坐标{    double a=B.x-A.x;    double b=B.y-A.y;    double c=D.x-C.x;    double d=D.y-C.y;    double e=A.x*B.y-A.y*B.x;    double f=C.x*D.y-C.y*D.x;    double y=(b*f-d*e)/(a*d-b*c);    double x=(a*f-c*e)/(a*d-b*c);    return Point(x,y);}double cross(Point a,Point b)//叉乘计算结果{    return a.x*b.y-a.y*b.x;}int main(){    sf("%d",&t);    while(t--)    {        double area=0;        sf("%lf%lf%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y,&D.x,&D.y);        F.x=A.x/3*2+B.x/3,F.y=A.y/3*2+B.y/3;//根据向量关系求坐标        D.x=B.x/3*2+C.x/3,D.y=B.y/3*2+C.y/3;        E.x=A.x/3+C.x/3*2,E.y=A.y/3+C.y/3*2;        X=getpoint(C,F,A,D);        Y=getpoint(B,E,A,D);        Z=getpoint(C,F,B,E);        area=cross(Y,X)+cross(Z,Y)+cross(X,Z);计算叉乘和求面积        area=fabs(area);        area/=2;        pf("%.0f\n",area);    }    return 0;}

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