码迷,mamicode.com
首页 > 其他好文 > 详细

hdu 5206 Four Inages Strategy 计算几何

时间:2015-04-18 21:55:36      阅读:155      评论:0      收藏:0      [点我收藏+]

标签:

题目链接:HDU - 5206

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技术分享 (no more than 10000技术分享 ), 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技术分享 ,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技术分享 : ans技术分享 , here x技术分享 is the data number begins at 1技术分享 , if your answer is yes,ans技术分享 is Yes, otherwise ans技术分享 is No.
题意描述:给出四个三维坐标位置,判断这四个点能否构成一个正方形。
算法分析:比赛的时候刚开始想到正方形边边相等,然后四角90度,这样的方法搞到一半的时候想了想有点麻烦呀,后来仔细想了想,突然意识到正方形里每个顶点到其他三个顶点距离的关系:对角线距离*对角线距离=2*边*边。然后就OK啦。
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<queue>
 9 #define inf 0x7fffffff
10 using namespace std;
11 typedef long long LL;
12 const LL maxn=10000+10;
13 
14 LL n,m;
15 struct Point
16 {
17     LL x,y,z;
18     Point (LL x=0,LL y=0,LL z=0):x(x),y(y),z(z){}
19 }a,b,c,d;
20 typedef Point Vector;
21 
22 double Dot(Vector A,Vector B) {return A.x*B.x + A.y*B.y; }
23 double Length(Vector A) {return sqrt(Dot(A,A)); }
24 double angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B)); }
25 
26 LL dis(Point A,Point B)
27 {
28     LL xx=(A.x-B.x)*(A.x-B.x);
29     LL yy=(A.y-B.y)*(A.y-B.y);
30     LL zz=(A.z-B.z)*(A.z-B.z);
31     return xx+yy+zz;
32 }
33 
34 int main()
35 {
36     int t,ncase=1;
37     scanf("%d",&t);
38     while (t--)
39     {
40         scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d",
41             &a.x,&a.y,&a.z,&b.x,&b.y,&b.z,&c.x,&c.y,&c.z,&d.x,&d.y,&d.z);
42         //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);
43         int flag=0;
44         LL len,len2,len3;
45         len=dis(a,b) ;len2=dis(a,c) ;len3=dis(a,d) ;
46         if ((len==len2&&len3==len+len2)||(len==len3&&len2==len+len3)||(len2==len3&&len==len2+len3))
47             flag++;
48         len=dis(b,a) ;len2=dis(b,c) ;len3=dis(b,d) ;
49         if ((len==len2&&len3==len+len2)||(len==len3&&len2==len+len3)||(len2==len3&&len==len2+len3))
50             flag++;
51             len=dis(c,a) ;len2=dis(c,b) ;len3=dis(c,d) ;
52         if ((len==len2&&len3==len+len2)||(len==len3&&len2==len+len3)||(len2==len3&&len==len2+len3))
53             flag++;
54             len=dis(d,a) ;len2=dis(d,b) ;len3=dis(d,c) ;
55         if ((len==len2&&len3==len+len2)||(len==len3&&len2==len+len3)||(len2==len3&&len==len2+len3))
56             flag++;
57         if (flag==4) printf("Case #%d: Yes\n",ncase++);
58         else printf("Case #%d: No\n",ncase++);
59     }
60     return 0;
61 }

 

 
 

hdu 5206 Four Inages Strategy 计算几何

标签:

原文地址:http://www.cnblogs.com/huangxf/p/4438115.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!