Four Inages Strateg

Four Inages Strategy

 

 Accepts: 186

 

 Submissions: 1374

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

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

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题目意思：判断四个点能否构成正方形

分析：我用的方法比较挫==几乎用了枚举，学长说可以找中点，判断四个点到这个点的距离相等，并且两两边应该有四个小的相等，两个大的相等==这是学长的思路，思路也很清晰，下面只贴上自己比较挫的。

实现代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
using namespace std;
int x1,y1,z1;
int x2,y2,z2;
int x3,y3,z3;
int x4,y4,z4;
int T;
int main()
{
    scanf("%d",&T);
    for(int i=1;i<=T;i++)
    {
        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);
        if((x2-x1)*(x3-x1)+(y2-y1)*(y3-y1)+(z2-z1)*(z3-z1)==0&&((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)+(z2-z1)*(z2-z1)==(x3-x1)*(x3-x1)+(y3-y1)*(y3-y1)+(z3-z1)*(z3-z1))&&(x4==x3+x2-x1&&y4==y3+y2-y1&&z4==z3+z2-z1))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x2-x1)*(x4-x1)+(y2-y1)*(y4-y1)+(z2-z1)*(z4-z1)==0&&((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)+(z2-z1)*(z2-z1)==(x4-x1)*(x4-x1)+(y4-y1)*(y4-y1)+(z4-z1)*(z4-z1))&&(x3==x4+x2-x1&&y3==y4+y2-y1&&z3==z4+z2-z1))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x3-x1)*(x4-x1)+(y3-y1)*(y4-y1)+(z3-z1)*(z4-z1)==0&&((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1)+(z3-z1)*(z3-z1)==(x4-x1)*(x4-x1)+(y4-y1)*(y4-y1)+(z4-z1)*(z4-z1))&&(x2==x4+x3-x1&&y2==y4+y3-y1&&z2==z4+z3-z1))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x1-x2)*(x3-x2)+(y1-y2)*(y3-y2)+(z1-z2)*(z3-z1)==0&&((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2)==(x3-x2)*(x3-x2)+(y3-y2)*(y3-y2)+(z3-z2)*(z3-z2))&&(x4==x3+x1-x2&&y4==y3+y1-y2&&z4==z3+z1-z2))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x1-x2)*(x4-x2)+(y1-y2)*(y4-y2)+(z1-z2)*(z4-z1)==0&&((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2)==(x4-x2)*(x4-x2)+(y4-y2)*(y4-y2)+(z4-z2)*(z4-z2))&&(x3==x4+x1-x2&&y3==y4+y1-y2&&z3==z4+z1-z2))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x3-x2)*(x4-x2)+(y3-y2)*(y4-y2)+(z3-z2)*(z4-z1)==0&&((x3-x2)*(x3-x2)+(y3-y2)*(y3-y2)+(z3-z2)*(z3-z2)==(x4-x2)*(x4-x2)+(y4-y2)*(y4-y2)+(z4-z2)*(z4-z2))&&(x1==x4+x3-x2&&y1==y4+y3-y2&&z1==z4+z3-z2))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x1-x3)*(x2-x3)+(y1-y3)*(y2-y3)+(z1-z3)*(z2-z3)==0&&((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3)+(z1-z3)*(z1-z3)==(x2-x3)*(x2-x3)+(y2-y3)*(y2-y3)+(z2-z3)*(z2-z3))&&(x4==x2+x1-x3&&y4==y2+y1-y3&&z4==z2+z1-z3))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x1-x3)*(x4-x3)+(y1-y3)*(y4-y3)+(z1-z3)*(z4-z3)==0&&((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3)+(z1-z3)*(z1-z3)==(x4-x3)*(x4-x3)+(y4-y3)*(y4-y3)+(z4-z3)*(z4-z3))&&(x2==x4+x1-x3&&y2==y4+y1-y3&&z2==z4+z1-z3))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x2-x3)*(x4-x3)+(y2-y3)*(y4-y3)+(z2-z3)*(z4-z3)==0&&((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3)+(z2-z3)*(z2-z3)==(x4-x3)*(x4-x3)+(y4-y3)*(y4-y3)+(z4-z3)*(z4-z3))&&(x1==x4+x2-x3&&y1==y4+y2-y3&&z1==z4+z2-z3))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x1-x4)*(x2-x4)+(y1-y4)*(y2-y4)+(z1-z4)*(z2-z4)==0&&((x1-x4)*(x1-x4)+(y1-y4)*(y1-y4)+(z1-z4)*(z1-z4)==(x2-x4)*(x2-x4)+(y2-y4)*(y2-y4)+(z2-z4)*(z2-z4))&&(x3==x2+x1-x4&&y3==y2+y1-y4&&z3==z2+z1-z4))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x1-x4)*(x3-x4)+(y1-y4)*(y3-y4)+(z1-z4)*(z3-z4)==0&&((x1-x4)*(x1-x4)+(y1-y4)*(y1-y4)+(z1-z4)*(z1-z4)==(x3-x4)*(x3-x4)+(y3-y4)*(y3-y4)+(z3-z4)*(z3-z4))&&(x2==x4+x1-x4&&y2==y4+y1-y4&&z2==z4+z1-z4))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else  if((x2-x4)*(x3-x4)+(y2-y4)*(y3-y4)+(z2-z4)*(z3-z4)==0&&((x2-x4)*(x2-x4)+(y2-y4)*(y2-y4)+(z2-z4)*(z2-z4)==(x3-x4)*(x3-x4)+(y3-y4)*(y3-y4)+(z3-z4)*(z3-z4))&&(x1==x3+x2-x4&&y1==y3+y2-y4&&z1==z3+z2-z4))
        {
            printf("Case #%d: Yes\n",i);
            continue;
        }
        else
        {
            printf("Case #%d: No\n",i);
            continue;
        }
    }
}

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