Four Inages Strategy hdu 5206

Four Inages Strategy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 182    Accepted Submission(s): 66

Problem Description

Young F found a secret record which inherited from ancient times in ancestral home by accident, which named "Four Inages Strategy". He couldn't restrain inner exciting, open the record, and read it carefully. " Place four magic stones at four points as array element in space, if four magic stones form a square, then strategy activates, destroying enemy around". Young F traveled to all corners of the country, and have collected four magic stones finally. He placed four magic stones at four points, but didn't know whether strategy could active successfully. So, could you help him?

 

Input

Multiple test cases, the first line contains an integer T(no more than 10000), indicating the number of cases. Each test case contains twelve integers x1,y1,z1,x2,y2,z2,x3,y3,z3,x4,y4,z4,|x|,|y|,|z|≤100000,representing coordinate of four points. Any pair of points are distinct.

 

Output

For each case, the output should occupies exactly one line. The output format is Case #x: ans, here x is the data number begins at 1, if your answer is yes,ans is Yes, otherwise ans is No.

 

Sample Input

2
0 0 0 0 1 0 1 0 0 1 1 0
1 1 1 2 2 2 3 3 3 4 4 4

 

Sample Output

Case #1: Yes
Case #2: No

就是求空间上的四个点看是否围成一个正方形。
先找邻边相等  再求对角线平方和为邻边的两倍~~

#include<stdio.h>
#include<string.h>
int main()
{
	int bbs;
	int x1,y1,z1,x2,y2,z2,x3,y3,z3,x4,y4,z4;
	scanf("%d",&bbs);
	int k=1;
	while(bbs--)
	{
		scanf("%d%d%d%d%d%d%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2,&x3,&y3,&z3,&x4,&y4,&z4);
		int sum1=(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)+(z2-z1)*(z2-z1);
		int sum2=(x2-x3)*(x2-x3)+(y2-y3)*(y2-y3)+(z2-z3)*(z2-z3);
		int sum3=(x2-x4)*(x2-x4)+(y2-y4)*(y2-y4)+(z2-z4)*(z2-z4);
		
		int a,b;
		if(sum1==sum2||sum2==sum3||sum3==sum1)
		{
			if(sum1==sum2)
			{
				a=sum1;
				b=sum3;
			}
			else if(sum2==sum3)
			{
				a=sum2;
				b=sum1;
			}
			else 
			{
				a=sum3;
				b=sum2;
			}
			if(a*2==b)
			{
				printf("Case #%d: Yes\n",k++);
			}
			else printf("Case #%d: No\n",k++);
		}
		else 
		{
			printf("Case #%d: No\n",k++);
		}		
	}
	return 0;
}