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Problem D: Tetrahedron Inequality

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 15   Solved: 3
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Description

It is well known that you cannot make a triangle with non-zero area whose sides have lengths 1, 2, 3. Can you make a tetrahedron(四面体) with non-zero volume whose edges have lengths 1, 2, 3, 4, 5, 6?

Input

 The first line of input contains an integer 0 < n <= 10000, the number of lines to follow. Each of the next n lines contains six positive integers separated by spaces, the lengths of the edges of the desired tetrahedron. The length of each edge is no greater than 1000000.

Output

Output n lines, each containing the word YES if it is possible to construct a tetrahedron with non-zero volume with the given edge lengths, or the word NO if it is not possible.

Sample Input

21 2 3 4 5 610 10 10 10 10 18

Sample Output

NONO

队友所写，之后，为这题不知争了多少次，调了不知道多长时间，怎么测都对，提交就是不对，最后原来这题

还是那个精度问题，因为，有可能是1000000,这样肯定，越界了的，所以必须太大的时候，除个１０００才行，警记

#include<stdio.h> 

#include<math.h> 

  

int judge(double a,double b,double c) //判定能否组成三角形

{ 

    if(a+b>c&&a+c>b&&b+c>a) 

        return 1; 

    else return 0; 

} 

  

int cal(double a,double b,double c,double d,double e,double f) 

{ 

    if(a>1000) 

    { 

        a/=1000.0; 

        b/=1000.0; 

        c/=1000.0; 

        d/=1000.0; 

        e/=1000.0; 

        f/=1000.0; 

    } 

    double L1,L2,h1,h2,x1,x2,ff; 

    L1=(a+b+d)/2.0; 

    L2=(a+c+e)/2.0; 

    h1=2*sqrt(L1)*sqrt(L1-a)/a*sqrt(L1-b)*sqrt(L1-d);// 

    h2=2*sqrt(L2)*sqrt(L2-a)/a*sqrt(L2-c)*sqrt(L2-e); 

    x1=(b*b+e*e-d*d-c*c)/4/a/a*(b*b+e*e-d*d-c*c);//是平方过的, 

    //x2=(b*b+e*e-d*d-c*c)*(b*b+e*e-d*d-c*c)/4/a/a; 

    ff=f*f; 

    if(ff<(h1+h2)*(h1+h2)+x1&&ff>(h1-h2)*(h1-h2)+x1) 

        return 1; 

    else return 0; 

} 

int main() 

{ 

    int t,a,b,c,d,e,f,i,j; 

    double r[6]; 

    scanf("%d",&t); 

    while(t--) 

    { 

       

       

        for(i=0;i<6;i++) 

            scanf("%lf",&r[i]); 

        for(j=0,a=0;a<5;a++) 

        { 

            for(b=0;b<5;b++) 

            { 

                if(a!=b) 

                for(c=0;c<5;c++) 

                {if(a!=c&&b!=c) 

                    for(d=0;d<5;d++) 

                    {if(a!=d&&d!=c&&b!=d&&judge(r[a],r[b],r[d])) 

                        for(e=0;e<5;e++) 

                        {if(a!=e&&d!=e&&b!=e&&c!=e) 

                            if(judge(r[a],r[c],r[e])) 

                            { 

                                if(cal(r[a],r[b],r[c],r[d],r[e],r[5])) 

                                { 

                                    j++; 

                                    break; 

                                } 

                            } 

                            if(j) 

                                break; 

                        } 

                        if(j) 

                            break; 

                    } 

                    if(j) 

                        break; 

                } 

                if(j) 

                    break; 

            } 

            if(j) 

                break; 

        } 

        if(j) 

            printf("YES\n"); 

        else printf("NO\n"); 

    } 

    return 0; 

}