如果一个人能统治无穷远处，那么他的速度一定是最大的。除了那几种很坑的数据，比如同一个点速度相同的两个人。永远都是不可能。所以你要处理出来所有速度最大的点，然后用他们构成一个凸包，你把端点的点求出来了，还得判断一下在边上的情况。然后顶点和在边上的点所构成的就是可以到达无穷远处的人。
PS：抄了芳姐的模版。。哈哈哈
Area of Mushroom
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 676 Accepted Submission(s): 120
Problem Description
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.
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).
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
Source
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define clearall(A, X) memset(A, X, sizeof(A))
using namespace std;
const int maxn = 2010;
struct point
{
int x,y;
int v,id;
point(int x = 0,int y = 0):x(x),y(y) {}
} p[maxn],ch[maxn],q[maxn],pq[maxn];
int f[maxn];
typedef point pointt;
point operator -(point a,point b)
{
return point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
if(fabs(x)<eps) return 0;
return x<0?-1:1;
}
int cross(point a,point b)
{
return a.x*b.y-a.y*b.x;
}
int mul(point a,point b,point c)
{
return cross(b-a,c-a);
}
double dis(point a)
{
return sqrt(a.x*a.x+a.y*a.y);
}
bool cmp(point a,point b)
{
if(mul(p[0],a,b)==0)
return dis(a-p[0])<dis(b-p[0]);
return mul(p[0],a,b)>0;
}
int graham(int n)///返回点的个数
{
int i,k =0 ,top;
if(n<2) return 0;
point tmp ;
for(i = 0; i < n ; i++)
{
if(dcmp(p[i].y-p[k].y)<0||(dcmp(p[i].y-p[k].y)==0&&dcmp(p[i].x-p[k].x)<0))
k = i;
}
swap(p[k],p[0]);
sort(p+1,p+n,cmp);
ch[0] = p[0];
ch[1] = p[1];
top = 1;
for(i = 2; i < n ; i++)
{
while(top>0&&dcmp(mul(ch[top-1],ch[top],p[i]))<=0) top--;
top++;
ch[top] = p[i];
}
return top;
}
bool cmpp(point a,point b)
{
if(a.x==b.x) return a.y<b.y;
return a.x<b.x;
}
int main()
{
int Case = 1;
int n;
while(cin >>n)
{
if(!n) break;
memset(f, 0, sizeof(f));
int Max = -INF;
for(int i = 1; i <= n; i++)
{
scanf("%d %d %d",&q[i].x, &q[i].y, &q[i].v);
q[i].id = i;
if(!q[i].v) f[i] = -2;
Max = max(Max, q[i].v);
}
for(int i = 1; i <= n; i++)
{
if(q[i].v != Max) continue;
for(int j = i+1; j <= n; j++)
{
if(q[j].x == q[i].x && q[j].y == q[i].y && q[j].v == q[i].v)
{
f[q[j].id] = -1;
f[q[i].id] = -1;
}
}
}
int t = 0;
for(int i = 1; i <= n; i++)
if(q[i].v == Max && f[q[i].id] != -2) p[t++] = q[i];
int m = graham(t);
cout<<"Case #"<<Case++<<": ";
if(m < 2)
{
for(int i = 1; i <= n; i++)
{
if(f[i] != -1 && f[i] != -2 && q[i].v == Max) cout<<1;
else cout<<0;
}
cout<<endl;
continue;
}
ch[m+1] = ch[0];
for(int i = 0; i <= m; i++) if(!f[ch[i].id]) f[ch[i].id] = 1;
for(int i = 1; i <= n; i++)
{
if(f[i] || q[i].v != Max) continue;
for(int j = 0; j <= m; j++)
if(mul(ch[j], ch[j+1], q[i]) == 0) f[i] = 1;
}
for(int i = 1; i <= n; i++)
{
if(f[i] == 1) cout<<1;
else cout<<0;
}
cout<<endl;
}
return 0;
}