HDU 3756 Dome of Circus （三分）

傳送門

題意很簡單，就是給你N個三維座標，要你求一個三菱錐把所有點罩進去，在邊邊也行，要你求出滿足條件的最小體積的三菱錐的高H和底面半徑R

很簡單三分就行了

但是這題我自己當時過得很慘烈，我寫搓了，結果靠交來測答案。

先是我寫搓的代碼，給巨巨們引以爲戒

#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--){//這裏我先三分高，再二分最合適的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;//這裏我通過作死交猜測對的
            size2 = size;//多一些會超時，少一些會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，之後我看了別人代碼後才知道，我是多麼幸運

#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);//這裏初始的左邊界就要做處理了，不然會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~