Contest - 2013校赛暨华中邀请赛 Problem D: Tetrahedron Inequality

Problem D: Tetrahedron Inequality

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 14 Solved: 2
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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

题意：输入六个数让你判断他们能否组成四面体。上次去抵达比赛没做出来的，当时想的太简单了，一直WA，比赛结束后才知道为什么，是任选五条边的各种三角形的组合没考虑。今天又做了一下，很快写好了，可是想AC还是不容易啊，具体的看注释吧

/**************************************************************

Problem: 1432

User: wust_*******

Language: C

Result: Accepted

Time:44 ms

Memory:760 kb

****************************************************************/

#include<stdio.h> #include<math.h>   int judge(double a,double b,double c) {     if(a+b>c&&a+c>b&&b+c>a) //能否组成三角形        return 1;     else return 0; }   int cal(double a,double b,double c,double d,double e,double f) {     if(a>1000) //因为题目说明了边的范围<=1000000，如果输入1000000的边长,看看后面的带乘法的运算每一个都能爆掉，这就是刚才WA好多的原因，所以对大数据要同时除以1000.0

    {         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);// 算高的    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);//两个三角形高之间的距离平方    ff=f*f;     if(ff<(h1+h2)*(h1+h2)+x1&&ff>(h1-h2)*(h1-h2)+x1) //根据任意可以组成两个三角形的五条边来算第六条边f的长度范围，看f是否在这个范围内        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) //保证不会选重边                    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])) //计算能否组成四面体                                {                                     j++;                                     break; //有一个成立就退出循环                                }                             }                             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; }