Area of Mushroom
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 806 Accepted Submission(s): 167
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
题解：m个点中，肯定速度最大的点才可以掌控无限的面积。然后这n个速度最大的点可以构成一个凸包，凸包内的点也坑爹不能掌控无限面积的。所以只需要求凸包上的点即可。注意，有两个点在同一位置他们是不能掌控无限面积的，但是却可以加入凸包中。还有我的凸包构成的时候，刚加进去的两个点也就是首尾两个点之间的点有可能被忽视掉。所以最后加个判断是否这两点之间存在点并加入。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#define LL __int64
#define ling 1e-10
using namespace std;
struct point
{
int x,y,pos,flag;
};
struct node
{
int x,y,z,pos,flag;
}f[510];
int ans[510];
point pp[510],p[510];
int sp,n;
double fff(point a,point b,point c)
{
return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
double dis(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int cmp(point a,point b)
{
double k=fff(p[0],a,b);
if (k<-ling)
return 0;
if (fabs(k)<=ling && dis(p[0],a)-dis(p[0],b)>ling)
return 0;
return 1;
}
int ff(int n)
{
int i,j=0;
point k=p[0];
for (i=1;i<n;i++)
{
if (k.y>p[i].y)
{
k=p[i];
j=i;
}
else
if (k.y==p[i].y && k.x>p[i].x)
{
k.x=p[i].x;
j=i;
}
}
point t;
t=p[0];
p[0]=p[j];
p[j]=t;
sort(p+1,p+n,cmp);
pp[0]=p[n-1];
pp[1]=p[0];
sp=1;
int l=1;
while (l<=n-1)
{
double d=fff(pp[sp],pp[sp-1],p[l]);
if( d < ling)
{
sp++;
pp[sp]=p[l];
l++;
}
else sp--;
}
}
int main()
{
int cas=1,i,m;
freopen("outt.txt","w",stdout);
while (~scanf("%d",&m) && m)
{
int max=-1;
for (i=1;i<=m;i++)
{
scanf("%d%d%d",&f[i].x,&f[i].y,&f[i].z);
f[i].pos=i;f[i].flag=0;ans[i]=0;
if (max<f[i].z) max=f[i].z;
}
printf("Case #%d: ",cas++);
if (max==0)
{
for (i=1;i<=m;i++)
printf("0");
cout<<endl;
continue;
}
n=0;
map<int,int>mp;
mp.clear();
memset(pp,0,sizeof(pp));
for (i=1;i<=m;i++)
if (max==f[i].z)
{
if (mp.find(f[i].x*10000+f[i].y) ==mp.end())
{
p[n].x=f[i].x;
p[n].y=f[i].y;
p[n].pos=f[i].pos;
p[n].flag=1;
mp[f[i].x*10000+f[i].y]=n;
//vis[f[i].x][f[i].y]=1;
n++;
}
else p[mp[f[i].x*10000+f[i].y]].flag=0;
}
if (n>3)
{
ff(n);
for (i=0;i<=sp;i++)
//if (fff(pp[i-1],pp[i],pp[i+1])==0 && i!=sp && i!=1) continue;
ans[pp[i].pos]=pp[i].flag;
n=n-2;
while (n>=0 && fff(pp[1],p[n],pp[0])==0)
{
ans[p[n].pos]=p[n].flag;
n--;
}
}
else
{
for (i=0;i<n;i++)
ans[p[i].pos]=p[i].flag;
}
for (i=1;i<=m;i++)
printf("%d",ans[i]);
cout<<endl;
}
return 0;
}