给出n个点,要你找到一个三角形,它的高是最长的。
思路:暴力超时了,是用先找出n个点与其他点的最长边,再枚举顶点过的.......具体证明不知道.....
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define eps 1e-8
struct point
{
double x;
double y;
};
//点到直线的最短距离
//bool vist[500][500][500];
point intersection(point u1,point u2,point v1,point v2)
{
point ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}
point ptoline(point p,point l1,point l2)
{
point t=p;
t.x+=l1.y-l2.y;
t.y+=l2.x-l1.x;
return intersection(p,t,l1,l2);
}
double juli(point a,point b)
{
return (sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)));
}
double str[505][2];
int main()
{
int n;
while(scanf("%d",&n)>0)
{
//memset(vist,false,sizeof(vist));
for(int i=0; i<n; i++)
scanf("%lf%lf",&str[i][0],&str[i][1]);
double maxn=0;
point a,b,c;
point sp[1000][2];
int cnt=0;
for(int i=0; i<n; i++)
{
a.x=str[i][0];
a.y=str[i][1];
double zd=0;
for(int j=0; j<n; j++)
{
if(i==j) continue;
b.x=str[j][0];
b.y=str[j][1];
double tmp=juli(a,b);
if(tmp>zd)
{
zd=tmp;
sp[cnt][0]=a;
sp[cnt][1]=b;
}
}
cnt++;
}
for(int i=0; i<n; i++)
{
a.x=str[i][0];
a.y=str[i][1];
for(int j=0; j<cnt; j++)
{
b=sp[j][0];
c=sp[j][1];
point d=ptoline(a,b,c);
maxn=max(maxn,juli(d,a));
d=ptoline(b,a,c);
maxn=max(maxn,juli(d,b));
d=ptoline(c,a,b);
maxn=max(maxn,juli(d,c));
}
}
printf("%.5lf\n",maxn);
}
return 0;
}