D - Area of Mushroom

Teacher Mai has a kingdom with the infinite area.

He has n students guarding the kingdom.

The i-th student stands at the position (x i,y i), and his walking speed is v i.

If a point can be reached by a student, and the time this student walking to this point is strictly less than other students, this point is in the charge of this student.

For every student, Teacher Mai wants to know if the area in the charge of him is infinite.

Input

There are multiple test cases, terminated by a line "0".

For each test case, the first line contains one integer n(1<=n<=500).

In following n lines, each line contains three integers x i,y i,v i(0<=|x i|,|y i|,v i<=10^4).

Output

For each case, output "Case #k: s", where k is the case number counting from 1, and s is a string consisting of n character. If the area in the charge of the i-th student isn't infinite, the i-th character is "0", else it's "1".

Sample Input

3
0 0 3
1 1 2
2 2 1
0

Sample Output

Case #1: 100

题意是有n我的，每一个人有一个坐标和速度，平面上若是的点若是他到达的时间严格的比其余任何人都快，那么这个点就属于他管辖。问每一个人的管辖区域是否是无穷大。显然对于两个速度不一样的人，速度小的人就不多是无穷大。因此只须要找出速度最大的全部的人。先求出凸包，凸包的顶点是无穷大，而后找到全部凸包边上的点，这些点也多是无穷大。而后暴力枚举全部的速度最大的点，若是和目前认为是无穷大的点重合，显然这两个重合点有相同速度到任何的点都是同样的时间，因此去掉。坑点是速度是0的点不可能无穷大。

wa了半天，我就是喜欢写代码bug

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<stack>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod=1e9+100;
const db e=exp(1);
const db eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int INF=0xfffffff;
struct Point{
    int x,y,num,temp1,temp2;
}p[1007],s[1007],a[1007];
int direction(Point p1,Point p2,Point p3) 
{ 
    return (p3.x-p1.x)*(p2.y-p1.y)-(p2.x-p1.x)*(p3.y-p1.y); 
}//点2和3，按哪一个和点一的角度更小排，相同的话按哪一个更近排 
double dis(Point p1,Point p2) { return sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y)); }
bool cmp(Point p1,Point p2)//极角排序 
{
    int temp=direction(p[0],p1,p2);
    if(temp<0)return true ;
    if(temp==0&&dis(p[0],p1)<dis(p[0],p2))return true;
    return false;
}
int Graham(int n)
{
    int top;
    int pos,minx,miny;
    minx=miny=INF;
    for(int i=0;i<n;i++)//找最下面的基点
        if(p[i].y<miny||(p[i].y==miny&&p[i].x<minx))
        {
            minx=p[i].x;
            miny=p[i].y;
            pos=i;
        }
    swap(p[0],p[pos]);
    sort(p+1,p+n,cmp);
    p[n]=p[0];
    s[0]=p[0];s[1]=p[1];s[2]=p[2];
    top=2;
    for(int i=3;i<=n;i++)
    {
        while(direction(s[top-1],s[top],p[i])>=0&&top>=2)top--;
        s[++top]=p[i] ;
    }
    return top;
}
int main()
{
    int n,ans=1;
    while(1)
    {
        sf("%d",&n);
        if(!n) return 0;
        int Max=0;
        rep(i,0,n)
        {
            a[i].temp1=0;a[i].temp2=0;
            sf("%d%d%d",&a[i].x,&a[i].y,&a[i].num);
            Max=max(Max,a[i].num);
        }
        if(Max==0)
        {
            pf("Case #%d: ",ans++);
            rep(i,0,n) 
            pf("0");
            pf("\n");
            continue;
        }
        rep(i,0,n) if(a[i].num==Max) a[i].temp1=1;
        rep(i,0,n)
        {
            if(a[i].num==Max)
            rep(j,0,n)
            {
                if(i!=j&&a[j].num==Max&&a[i].x==a[j].x&&a[i].y==a[j].y)
                {
                    a[i].temp1=0;a[j].temp1=0;
                }
            }
        } 
        int sum=0;
        rep(i,0,n)
        {
            if(a[i].num==Max)
            {
                int temp=0;
                rep(j,0,sum)
                    if(a[i].x==p[j].x&&a[i].y==p[j].y)
                        temp=1;
                if(temp==0)
                    p[sum++]=a[i];
            }
        }
        int top;
        top=Graham(sum);
        rep(i,0,top)
        {
            rep(j,0,n)
            {
                if(s[i].x==a[j].x&&s[i].y==a[j].y)
                a[j].temp2=1;
                if(direction(s[i],s[(i+1)%top],a[j])==0)
                a[j].temp2=1;
            }
        }
        pf("Case #%d: ",ans++);
        rep(i,0,n)
        if(a[i].temp1&&a[i].temp2)
        pf("1");
        else
        pf("0");
        pf("\n");
    }
}