POJ 1755 Triathlon 半平面交

题意

题意：在铁人三项比赛中，给出每个人游泳，骑车，跑步的速度，你可以任意安排游泳，骑车，跑步的路程（路程非0），若某人可以获得第一（严格），则输出Yes，否则No

思路

将三个路程其中一个设为1，剩下两个是x,y.

对于每个人i，列出方程Ai*x+Bi*y+Ci<Aj*x+Bj*x+Cj。求半平面交，若解集为空，No。否则Yes。

代码

#include <iostream>
#include <cmath>
#include <cstring>
#include <vector>
#include <cstdio>
using namespace std;
#define INF 12345678
#define EPS 1e-8
#define NMAX 110
struct point
{
double x,y;
bool mark;
point(){}
point(double x0,double y0):x(x0),y(y0),mark(1){}
};
vector<point> p;
int n;
double a[NMAX],b[NMAX],c[NMAX];
point operator -(const point &A,const point &B)
{
return point(A.x-B.x,A.y-B.y);
}
double cross(point p1,point p2)
{
return p1.x*p2.y-p1.y*p2.x;
}
int dblcmp(double x)
{
if(fabs(x)<EPS) return 0;
return (x>0) ? 1:-1;
}
point find_intersection(point p1,point p2,point p3,point p4)
{
double s1=cross(p2-p1,p3-p1);
double s2=cross(p4-p1,p2-p1);
return point((p3.x*s2+p4.x*s1)/(s1+s2),(p3.y*s2+p4.y*s1)/(s1+s2));
}
void update(double A,double B,double C)
{
point t;
for (int i=0;i<p.size();i++)
if (dblcmp(A*p[i].x+B*p[i].y+C) >0) p[i].mark=1;
else p[i].mark=0;
p.push_back(p[0]);
for (int i=0;i<p.size()-1;i++)
{
if (p[i].mark ^ p[i+1].mark==1)
{
if (A!=0)
t=find_intersection(point(-C/A,0),point(-(C+B)/A,1),p[i],p[i+1]);
else if(B!=0)
t=find_intersection(point(0,-C/B),point(1,-C/B),p[i],p[i+1]);
p.insert(p.begin()+i+1,t);
i++;
}
}
p.pop_back();
for(int i=0;i<p.size();i++)
if(!p[i].mark)
p.erase(p.begin()+i), i--;
}
bool solve(int v)
{
p.clear();
p.push_back(point(INF,INF));
p.push_back(point(INF,0));
p.push_back(point(0,0));
p.push_back(point(0,INF));
for (int i=0;i<n;i++)
if (i!=v)
{
update(1/a[i]-1/a[v],1/b[i]-1/b[v],1/c[i]-1/c[v]);
if(p.size()==0) return 0;
}
return 1;
}
int main()
{
scanf("%d",&n);
for (int i=0;i<n;i++)
scanf("%lf%lf%lf",a+i,b+i,c+i);
for (int i=0;i<n;i++)
if (solve(i)) printf("Yes\n");
else printf("No\n");
return 0;
}

链接：http://poj.org/problem?id=1755