Four Inages Strategy
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 331 Accepted Submission(s): 128
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
Source
题意：
给你四个三维坐标点(x1,y1,z1),判断这四个点能否组成一个正方形。
题解：
吧每两个点的边长先算出来，排序，判断。
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<map>
#include<cmath>
using namespace std;
int n;
long long x[4];
long long y[4];
long long z[4];
long long a[10];
int main() {
int t;
cin>>t;
int ca=1;
while(t--) {
for(int i=0; i<4; i++)
scanf("%I64d%I64d%I64d",&x[i],&y[i],&z[i]);
int l=0;
for(int i=0; i<4; i++) {
for(int j=i+1; j<4; j++) {
a[l++]=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])+(z[i]-z[j])*(z[i]-z[j]);
}
}
sort(a,a+l);
if(a[0]==a[1]&&a[1]==a[2]&&a[2]==a[3]&&a[4]==a[5]&&(a[1]+a[2]==a[5])) {
printf("Case #%d: Yes\n",ca++);
} else
printf("Case #%d: No\n",ca++);
}
}
hdu 5206 Four Inages Strategy
原文：http://blog.csdn.net/acm_baihuzi/article/details/45119559
