HDU 3756 Dome of Circus （三分）

传送门

题意很简单，就是给你N个三维坐标，要你求一个三菱锥把所有点罩进去，在边边也行，要你求出满足条件的最小体积的三菱锥的高H和底面半径R

很简单三分就行了

但是这题我自己当时过得很惨烈，我写搓了，结果靠交来测答案。

先是我写搓的代码，给巨巨们引以为戒

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL __int64
using namespace std;
const double PI = acos(-1.0);
const int size = 50;
struct point
{
double x,y,z;
};

point insect[110000];
int n;

double checkr(double h,double r)
{
for(int i=0;i<n;i++){
if(insect[i].z > h) return false;
double R = (h-insect[i].z)*r/h;
if((insect[i].x*insect[i].x + insect[i].y*insect[i].y) > R*R) return false;
}
return true;
}

double calc(double h,double r)
{
return h*r*r*PI/3.0;
}

int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%lf%lf%lf",&insect[i].x,&insect[i].y,&insect[i].z);
}
double l = 0 , r = 3000,ll,rr;
double ans1,ans2,ans3,ans4;
double mid,midmid,mid2,midmid2;

int size1 = size,size2;
while(size1--){//这里我先三分高，再二分最合适的R
mid = (r+l)/2;
midmid = (mid+r)/2;

ll = 0 , rr = 3000;
size2 = size;
while(size2--){
mid2 = (ll+rr)/2;

if(checkr(mid,mid2))
{
ans3 = mid2;
rr = mid2;
}else ll = mid2;

}
ll = 0 , rr = 3000;//这里我通过作死交猜测对的
size2 = size;//多一些会超时，少一些会WA
while(size2--){
mid2 = (ll+rr)/2;
if(checkr(midmid,mid2))
{
ans4 = mid2;
rr = mid2;
}else ll = mid2;
}
if(calc(mid,ans3) > calc(midmid,ans4)){
l = mid;
ans1 = midmid;
ans2 = ans4;
}else{
r = midmid;
}
}
printf("%.3f %.3f\n",ans1,ans2);
}
return 0;
}

然而，之后我看了别人代码后才知道，我是多么幸运

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL __int64
using namespace std;
const double PI = acos(-1.0);
const int size = 50;
struct point
{
double x,y,z;
};

point insect[110000];
int n;
double ans1,ans2;

double calc(double h,double r)
{
return h*r*r*PI/3.0;
}

double checkr(double h)
{
ans2 = 0;
for(int i=0;i<n;i++){
double x = sqrt(insect[i].x*insect[i].x + insect[i].y*insect[i].y);
double R = x*h/(h-insect[i].z);
ans2 = max(ans2,R);
}
return calc(h,ans2);
}

int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
double l = 0 , r = 3000;
for(int i=0;i<n;i++){
scanf("%lf%lf%lf",&insect[i].x,&insect[i].y,&insect[i].z);
l = max(l,insect[i].z);//这里初始的左边界就要做处理了，不然会WA
}

double mid,midmid;
while(fabs(r-l)>1e-9){
mid = l + (r-l)/3.0;
midmid = r - (r-l)/3.0;
if(checkr(mid) > checkr(midmid)){
l = mid;
ans1 = midmid;
}else r = midmid;
}
printf("%.3f %.3f\n",ans1,ans2);
}
return 0;
}

加油~