2019 ACM-ICPC 南京 現場賽 K. Triangle (二分)

題意

給定一個三角形和一個點 \(p\)，如果該點不在三角形邊上直接輸出 \(-1\)，否則在三角形上找一點 \(q\)，使得線段 \(pq\) 平分三角形面積。

思路

看完題想都沒想直接二分了。

就是比賽時復制粘貼的時候改錯了，貢獻了三發罰時。

板子來自 kuangbin 的模板。

#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);
	}
	//叉積
	db operator ^(const Point &b)const{
		return x*b.y - y*b.x;
	}
	//點積
	db operator *(const Point &b)const{
		return x*b.x + y*b.y;
	}
	//返回兩點的距離
	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;
	}
	// 點在線段上的判斷
	bool pointonseg(Point p){
		return sgn((p-s)^(e-s)) == 0 && sgn((p-s) * (p-e)) <= 0;
	}
	
	// 求兩直線的交點
	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));
	}
};

// 求點a和點b的中點
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);  // 三角形面積
        if(ab.pointonseg(p)) {  // 點p在線段ab上
            if(a.dis(p) < b.dis(p)) {  // 點p靠近點a，則另一點一定在線段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 {  // 點p靠近點b，則另一點一定在線段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)) {  // 另外兩種情況復制粘貼就行了
            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;
}

賽后發現直接推也可以，也不用寫這么長的代碼了 (雖然大部分是復制粘貼)。感覺在演隊友。