The meaning of problems
Given a triangle and a point \ (P \) , if the point is not directly output triangle edge \ (- 1 \) , or to find the triangular point \ (Q \) , such that the line segment \ (PQ \) bisects the triangle area.
Thinking
After reading the title even imagined direct-half.
It's time to copy and paste the mistakes in the game, he contributed three rounds of penalty time.
Board from kuangbin template.
#include <bits/stdc++.h>
using namespace std;
typedef double db;
const db eps = 1e-8;
const db inf = 1e20;
const db pi = acos(-1.0);
int sgn(db x){
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
struct Point{
db x, y;
Point(){}
Point(db _x, db _y){
x = _x;
y = _y;
}
void input(){
scanf("%lf%lf", &x, &y);
}
bool operator == (Point b)const{
return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
}
Point operator -(const Point &b)const{
return Point(x-b.x, y-b.y);
}
//叉积
db operator ^(const Point &b)const{
return x*b.y - y*b.x;
}
//点积
db operator *(const Point &b)const{
return x*b.x + y*b.y;
}
//返回两点的距离
db dis(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 db &k)const{
return Point(x*k, y*k);
}
Point operator /(const db &k)const{
return Point(x/k, y/k);
}
};
struct Line{
Point s,e;
Line(){}
Line(Point _s,Point _e){
s = _s;
e = _e;
}
// 点在线段上的判断
bool pointonseg(Point p){
return sgn((p-s)^(e-s)) == 0 && sgn((p-s) * (p-e)) <= 0;
}
// 求两直线的交点
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));
}
};
// 求点a和点b的中点
Point get_mid(Point a, Point b) {
return (a + b) * 0.5;
}
int main() {
int T;
scanf("%d", &T);
while(T--) {
Point a, b, c, p;
Line ab, bc, ac;
a.input(), b.input(), c.input(), p.input();
ab = Line(a, b); bc = Line(b, c); ac = Line(a, c);
db area = fabs((b - a) ^ (c - a) * 0.5); // 三角形面积
if(ab.pointonseg(p)) { // 点p在线段ab上
if(a.dis(p) < b.dis(p)) { // 点p靠近点a，则另一点一定在线段bc上
Point l = c, r = b;
Point mid = get_mid(l, r);
int times = 1000;
while(times--) {
mid = get_mid(l, r);
db s = fabs((mid - p) ^ (b - p));
if(sgn(s - area) > 0) {
l = mid;
} else if(sgn(s - area) == 0) {
break;
} else {
r = mid;
}
}
printf("%.10lf %.10lf\n", mid.x, mid.y);
} else { // 点p靠近点b，则另一点一定在线段ac上
Point l = c, r = a;
Point mid = get_mid(l, r);
int times = 1000;
while(times--) {
mid = get_mid(l, r);
db s = fabs((mid - p) ^ (a - p));
if(sgn(s - area) > 0) {
l = mid;
} else if(sgn(s - area) == 0) {
break;
} else {
r = mid;
}
}
printf("%.10lf %.10lf\n", mid.x, mid.y);
}
} else if(bc.pointonseg(p)) { // 另外两种情况复制粘贴就行了
if(b.dis(p) < c.dis(p)) {
Point l = a, r = c;
Point mid = get_mid(l, r);
int times = 1000;
while(times--) {
mid = get_mid(l, r);
db s = fabs((mid - p) ^ (c - p));
if(sgn(s - area) > 0) {
l = mid;
} else if(sgn(s - area) == 0) {
break;
} else {
r = mid;
}
}
printf("%.10lf %.10lf\n", mid.x, mid.y);
} else {
Point l = a, r = b;
Point mid = get_mid(l, r);
int times = 1000;
while(times--) {
mid = get_mid(l, r);
db s = fabs((mid - p) ^ (b - p));
if(sgn(s - area) > 0) {
l = mid;
} else if(sgn(s - area) == 0) {
break;
} else {
r = mid;
}
}
printf("%.10lf %.10lf\n", mid.x, mid.y);
}
} else if(ac.pointonseg(p)) {
if(a.dis(p) < c.dis(p)) {
Point l = b, r = c;
Point mid = get_mid(l, r);
int times = 1000;
while(times--) {
mid = get_mid(l, r);
db s = fabs((mid - p) ^ (c - p));
if(sgn(s - area) > 0) {
l = mid;
} else if(sgn(s - area) == 0) {
break;
} else {
r = mid;
}
}
printf("%.10lf %.10lf\n", mid.x, mid.y);
} else {
Point l = b, r = a;
Point mid = get_mid(l, r);
int times = 1000;
while(times--) {
mid = get_mid(l, r);
db s = fabs((mid - p) ^ (a - p));
if(sgn(s - area) > 0) {
l = mid;
} else if(sgn(s - area) == 0) {
break;
} else {
r = mid;
}
}
printf("%.10lf %.10lf\n", mid.x, mid.y);
}
} else {
printf("-1\n");
}
}
return 0;
}After the game can also be found directly push, do not write such a long code (Although most copy and paste)。In his teammates feel。