Triathlon
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 6461 | Accepted: 1643 |
Description
Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first section is swimming, the second section is riding bicycle and the third one is running.
The speed of each contestant in all three sections is known. The judge can choose the length of each section arbitrarily provided that no section has zero length. As a result sometimes she could choose their lengths in such a way that some particular contestant would win the competition.
Input
The first line of the input file contains integer number N (1 <= N <= 100), denoting the number of contestants. Then N lines follow, each line contains three integers Vi, Ui and Wi (1 <= Vi, Ui, Wi <= 10000), separated by spaces, denoting the speed of ith contestant in each section.
Output
For every contestant write to the output file one line, that contains word "Yes" if the judge could choose the lengths of the sections in such a way that this particular contestant would win (i.e. she is the only one who would come first), or word "No" if this is impossible.
Sample Input
9
10 2 6
10 7 3
5 6 7
3 2 7
6 2 6
3 5 7
8 4 6
10 4 2
1 8 7
Sample Output
Yes
Yes
Yes
No
No
No
Yes
No
Yes
/*
poj 1755 半平面交+不等式 一个比赛分三个部分,每个人在三个部分的速度为U,V,W。每个赛道的长度不一定。
现在给你n个人的情况,问他们是否能得奖 总时间 t1 = x/u1+y/v1+z/w1 t2 = x/u2+y/v2+z/w2
那么 两个人的时间差 t = t1 - t2 = ax+by+cz,判断正负即可
所以 可以看成 (a/z)x+(b/z)y+c 就成了二元方程
然后利用半平面相交计算出这些不等式最后能否得到一个>0公共区域。
如果能，则说明冠军与你有缘诶 hhh-2016-05-17 22:32:51
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
using namespace std;
const int maxn = 300;
const double PI = 3.1415926;
const double eps = 1e-16;
int n;
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;
}
Point operator -(const Point &b)const
{
return Point(x-b.x,y-b.y);
}
double operator ^(const Point &b)const
{
return x*b.y-y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
}; struct Line
{
Point s,t;
double k;
Line() {}
Line(Point _s,Point _t)
{
s = _s;
t = _t;
k = atan2(t.y-s.y,t.x-s.x);
}
Point operator &(const Line &b) const
{
Point res = s;
double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
res.x += (t.x-s.x)*ta;
res.y += (t.y-s.y)*ta;
return res;
}
}; //求p1,p2的直线与a,b,c这条直线的交点
Point Intersection(Point p1,Point p2,double a,double b,double c)
{
double u = fabs(a*p1.x + b*p1.y + c);
double v = fabs(a*p2.x + b*p2.y + c);
Point t;
t.x = (p1.x*v + p2.x*u)/(u+v);
t.y = (p1.y*v + p2.y*u)/(u+v);
return t;
} double CalArea(Point p[],int n)
{
double ans = 0;
for(int i = 0; i < n; i++)
{
ans += (p[i]^p[(i+1)%n])/2;
}
return fabs(ans);
} double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
} Point p[maxn];
Point tp[maxn];
void cut(double a,double b,double c,Point p[],int &cnt)
{
int tmp = 0;
for(int i = 1; i <= cnt; i++)
{
if(a*p[i].x+b*p[i].y+c < eps) tp[++tmp] = p[i];
else
{
//在p[i]处大于0,那么交点可能在(p[i-1],p[i])or(p[i+1],p[i])
if(a*p[i-1].x + b*p[i-1].y + c < -eps)
tp[++tmp] = Intersection(p[i-1],p[i],a,b,c);
if(a*p[i+1].x + b*p[i+1].y + c < -eps)
tp[++tmp] = Intersection(p[i],p[i+1],a,b,c);
}
}
for(int i = 1; i <= tmp; i++)
p[i] = tp[i];
p[0] = p[tmp];
p[tmp+1] = p[1];
cnt = tmp;
}
double inf = 1000000000000000.0;
double U[maxn],V[maxn],W[maxn];
bool cal(int now)
{
p[1] = Point(0,0);
p[2] = Point(0,inf);
p[3] = Point(inf,inf);
p[4] = Point(inf,0);
p[0] = p[4];
p[5] = p[1];
int cnt = 4;
for(int i = 0; i < n; i++)
{
if(i == now) continue;
double a = (U[i]-U[now])/(U[i]*U[now]); //1/U[now] - 1/U[i]
double b = (V[i]-V[now])/(V[i]*V[now]);
double c = (W[i]-W[now])/(W[i]*W[now]);
if(sgn(a)==0 && sgn(b) == 0 )
{
if(sgn(c) >= 0)
return false;
else
continue;
}
cut(a,b,c,p,cnt);
}
if(sgn(CalArea(p,cnt)) == 0)
return false;
else
return true;
} int main()
{
// freopen("in.txt","r",stdin);
while(scanf("%d",&n)!= EOF)
{
for(int i = 0; i < n; i++)
{
scanf("%lf%lf%lf",&U[i],&V[i],&W[i]);
}
for(int i = 0; i < n; i++)
{
if(cal(i))
printf("Yes\n");
else
printf("No\n");
}
}
return 0;
}