2019 ICPC 南京 K.Triangle(二分+几何)

题意：

给一个三角形，以及一个点(某条线段的端点)，让求出另一点，使得这个线段平分这个三角形即两部分面积相等。

思路：

读完题就觉得是二分，奈何没有板子敲了半天，面积还是算不出来，最后抄了一手板子一发过。
首先如果p点不在三角形上面，直接输出-1
如果在三角形上面，那么可以判断出另一点是在另外两条边的哪条边上，比如p点在ab上且离a更近，那么必然在bc上，反之在ac上。

代码:

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n,m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sc(n) scanf("%c",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)
#define mem(a,n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define mod(x) ((x)%MOD)
#define gcd(a,b) __gcd(a,b)
#define lowbit(x) (x&-x)
#define pii map<int,int>
#define mk make_pair
#define rtl rt<<1
#define rtr rt<<1|1typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
//const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
//const int inf = 0x3f3f3f3f;
inline int read()
{int ret = 0, sgn = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')sgn = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){ret = ret*10 + ch - '0';ch = getchar();}return ret*sgn;
}
inline void Out(int a)
{if(a>9) Out(a/10);putchar(a%10+'0');
}
int qpow(int m, int k, int mod)
{int res = 1, t = m;while (k){if (k&1)res = res * t % mod;t = t * t % mod;k >>= 1;}return res;
}
ll gcd(ll a,ll b){return b==0?a : gcd(b,a%b);}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll inv(ll x,ll m){return qpow(x,m-2,m)%m;}const int N = 5e5+10;
int n,m;
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;
}// 根据三个点计算三角形面积
double area(Point a,Point b,Point c)
{return fabs((b - a) ^ (c - a) * 0.5);
}signed main()
{int t = 1;Point a,b,c,p;Line ab,ac,bc;sd(t);while(t--){a.input(),b.input(),c.input(),p.input();ab = Line(a,b),ac = Line(a,c),bc = Line(b,c);db s2 = area(a,b,c)/2;if(!ab.pointonseg(p) && !ac.pointonseg(p) && !bc.pointonseg(p)){cout<<-1<<endl;continue;}if(ab.pointonseg(p)){// 离a点更近 那么另一点在bc上 反之在ac上。if(a.dis(p) < b.dis(p)){Point l = b,r = c;Point mid = get_mid(l,r);int time = 1000;while(time--){mid = get_mid(l,r);db s = area(mid,p,b);int flag = sgn(s-s2);if(flag == 0)break;if(flag == 1)r = mid;elsel = mid;}printf("%.10lf %.10lf\n",mid.x,mid.y);}else{Point l = a,r = c;Point mid = get_mid(l,r);int time = 1000;while(time--){mid = get_mid(l,r);db s = area(mid,p,a);int flag = sgn(s-s2);if(flag == 0)break;if(flag == 1)r = mid;elsel = mid;}printf("%.10lf %.10lf\n",mid.x,mid.y);}}else if(ac.pointonseg(p)){if(a.dis(p) < c.dis(p)){Point l = c,r = b;Point mid = get_mid(l,r);int time = 1000;while(time--){mid = get_mid(l,r);db s = area(mid,p,c);int flag = sgn(s-s2);if(flag == 0)break;if(flag == 1)r = mid;elsel = mid;}printf("%.10lf %.10lf\n",mid.x,mid.y);}else{Point l = a,r = b;Point mid = get_mid(l,r);int time = 1000;while(time--){mid = get_mid(l,r);db s = area(mid,p,a);int flag = sgn(s-s2);if(flag == 0)break;if(flag == 1)r = mid;elsel = mid;}printf("%.10lf %.10lf\n",mid.x,mid.y);}}else{if(b.dis(p) < c.dis(p)){Point l = c,r = a;Point mid = get_mid(l,r);int time = 1000;while(time--){mid = get_mid(l,r);db s = area(mid,p,c);int flag = sgn(s-s2);if(flag == 0)break;if(flag == 1)r = mid;elsel = mid;}printf("%.10lf %.10lf\n",mid.x,mid.y);}else{Point l = b,r = a;Point mid = get_mid(l,r);int time = 1000;while(time--){mid = get_mid(l,r);db s = area(mid,p,b);int flag = sgn(s-s2);if(flag == 0)break;if(flag == 1)r = mid;elsel = mid;}printf("%.10lf %.10lf\n",mid.x,mid.y);}}}return 0;
}

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2019 ICPC 南京 K.Triangle(二分+几何)

题意：

思路：

代码:

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