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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; 
}