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

題意是有n我的，每一個人有一個座標和速度，平面上若是的點若是他到達的時間嚴格的比其餘任何人都快，那麼這個點就屬於他管轄。問每一個人的管轄區域是否是無窮大。顯然對於兩個速度不一樣的人，速度小的人就不多是無窮大。因此只須要找出速度最大的全部的人。先求出凸包，凸包的頂點是無窮大，而後找到全部凸包邊上的點，這些點也多是無窮大。而後暴力枚舉全部的速度最大的點，若是和目前認爲是無窮大的點重合，顯然這兩個重合點有相同速度到任何的點都是同樣的時間，因此去掉。坑點是速度是0的點不可能無窮大。

wa了半天，我就是喜歡寫代碼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); 
}//點2和3，按哪一個和點一的角度更小排，相同的話按哪一個更近排 
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)//極角排序 
{
    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++)//找最下面的基點
        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");
    }
}