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題目大意：給出三個頂點A，B，C組成三角形，如題目中的圖片所示，規定D，E，F分別為三條邊的三等分點，現在要求三角形RPQ的面積

題目分析：利用向量先求出DEF三點，再利用線段相交求出RPQ三點，最后利用海倫公式直接求答案，還是考察了模板的配合

代碼：
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#include<iostream> #include<cstdio> #include<string> #include<ctime> #include<cmath> #include<cstring> #include<algorithm> #include<stack> #include<queue> #include<map> #include<set> #include<sstream> using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const int N=2e5+100; //字符串長度最大值const double eps = 1e-8;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;}void input(){scanf("%lf%lf",&x,&y);}//返回兩點的距離double distance(Point p){return hypot(x-p.x,y-p.y);}Point operator +(const Point &b)const{return Point(x+b.x,y+b.y);}Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}//叉積double operator ^(const Point &b)const{return x*b.y - y*b.x;}//點積double operator *(const Point &b)const{return x*b.x + y*b.y;}Point operator /(const double &k)const{return Point(x/k,y/k);} };struct Line{Point s,e;Line(){}Line(Point _s,Point _e){s = _s;e = _e;}//`兩線段相交判斷`//`2 規范相交`//`1 非規范相交`//`0 不相交`int segcrossseg(Line v){int d1 = sgn((e-s)^(v.s-s));int d2 = sgn((e-s)^(v.e-s));int d3 = sgn((v.e-v.s)^(s-v.s));int d4 = sgn((v.e-v.s)^(e-v.s));if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||(d4==0 && sgn((e-v.s)*(e-v.e))<=0);}//`求兩直線的交點`//`要保證兩直線不平行或重合`Point crosspoint(Line v){double a1 = (v.e-v.s)^(s-v.s);double a2 = (v.e-v.s)^(e-v.s);return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));} };Point get_point(Point a,Point b) {return a+(b-a)/3; }Point get_point(Line a,Line b) {return a.crosspoint(b); }double area(Point a,Point b,Point c) {double lena=a.distance(b);double lenb=a.distance(c);double lenc=b.distance(c);double p=(lena+lenb+lenc)/2;return sqrt(p*(p-lena)*(p-lenb)*(p-lenc)); }int main() { // freopen("input.txt","r",stdin); // ios::sync_with_stdio(false);int w;cin>>w;while(w--){Point A,B,C;A.input(),B.input(),C.input();Point D,E,F;D=get_point(B,C);E=get_point(C,A);F=get_point(A,B);Point P,Q,R;P=get_point(Line(B,E),Line(A,D));Q=get_point(Line(C,F),Line(B,E));R=get_point(Line(C,F),Line(A,D));printf("%.0f\n",area(P,Q,R));}return 0; }

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超強干貨來襲 云風專訪：近40年碼齡，通宵達旦的技術人生

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