POJ 1755 Triathlon (计算几何+半平面交解决线性规划)

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

分别对每个人列一组不等式方程组再分别对每一组不等式方程组用半平面交求解

因为没规定跑道长度，我一开始就自己定义了一个很长的长度L，然后三个项目每一段长度为x,y,(L-x-y),

然后利用每一段的速度来求出每个人的总时间以此来列不等式方程组，但是这样wa了N发。。。

我们可以直接定义三段的长度为x,y,1( 可以看成以一定比例缩小之后，这样就稳了 )，然后列式子，半平面交

代码：

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<functional>
#include<vector>
#include <map>
#include<cmath>
using namespace std;
typedef long long LL;
const LL MOD=1e9+7;
const double eps=1e-9;
const int MAXN=100+23;
#define point pair<double,double>
#define x first
#define y second
void output(point p[],int n)
{
    puts("******points:");
    for(int i=1;i<=n;i++)
    {
        printf("%.0f %.0f\n",p[i].x,p[i].y);
    }
}
double cross(point o,point a,point b)
{
    return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
double getarea(point p[],int n)
{
    double area=0;
    for(int i=2;i<n;i++)
    {
        area+=(p[i].x-p[1].x)*(p[i+1].y-p[1].y)-(p[i].y-p[1].y)*(p[i+1].x-p[1].x);
    }
    return area/2.0;
}
double dist(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void getline(double &a,double &b,double &c,point x1,point x2)
{
    a=(x1.y-x2.y);
    b=(x2.x-x1.x);
    c=-a*x1.x-b*x1.y;
}
point intersect(double a,double b,double c,point x1,point x2)
{
    double u=fabs(a*x1.x+b*x1.y+c);
    double v=fabs(a*x2.x+b*x2.y+c);
    return point((x1.x*v+x2.x*u)/(u+v),(x1.y*v+x2.y*u)/(u+v));
}
int n,last,now;
point p1[MAXN],p2[MAXN],p3[MAXN];
void cut(double a,double b,double c)
{
    now=0;
    //if(a==0&&b==0&&c<eps){last=0;return;}
    for(int i=1;i<=last;i++)
    {
        if((a*p2[i].x+b*p2[i].y+c)>eps)
        {
            p3[++now]=p2[i];
        }
        else
        {
            if((a*p2[i-1].x+b*p2[i-1].y+c)>eps)
            {
                p3[++now]=intersect(a,b,c,p2[i],p2[i-1]);
            }
            if((a*p2[i+1].x+b*p2[i+1].y+c)>eps)
            {
                p3[++now]=intersect(a,b,c,p2[i],p2[i+1]);
            }
        }
    }
    last=now;
    for(int i=1;i<=last;i++)p2[i]=p3[i];
    p2[last+1]=p3[1];p2[0]=p3[last];
}
inline void GuiZhengHua()
{
    //规整化方向，逆时针变顺时针，顺时针变逆时针
    for(int i = 1; i <=(n)/2; i++)
      swap(p1[i], p1[n-i]);//头文件加iostream
}
double length=1<<28;
double k1[100+23],k2[100+23],k3[100+23];
int main()
{
    while(scanf("%d",&n)!=-1)
    {
        for(int i=1;i<=n;i++)
        {
            double u,v,w;
            scanf("%lf%lf%lf",&u,&v,&w);
            k3[i]=w;
            k1[i]=u;
            k2[i]=v;
        }
        int m=n;
        for(int i=1;i<=m;i++)
        {
            p2[1]=point(0,0);
            p2[2]=point(0,length);
            p2[3]=point(length,length);
            p2[4]=point(length,0);
            p2[0]=p2[4];
            p2[5]=p2[1];
            last=4;
            double a,b,c;
            for(int j=1;j<=m;j++)
            {
                if(j==i)continue;
                a=(k1[i]-k1[j])/(k1[i]*k1[j]),b=(k2[i]-k2[j])/(k2[i]*k2[j]),c=(k3[i]-k3[j])/(k3[i]*k3[j]);
                cut(a,b,c);
                if(last==0)break;
            }
            if(getarea(p2,last)>=0)puts("No");
            else puts("Yes");
        }
    }
    return 0;
}

本人蒟蒻，如有错误，还望指正