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题目链接:
Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=210;
const double eps=1e-12;
int n;
struct PO
{
int x,y,z;
}po[maxn];
int check1(PO a,PO b,PO c)//判断共线;
{
PO temp1,temp2;
temp1.x=a.x-b.x;temp1.y=a.y-b.y;temp1.z=a.z-b.z;
temp2.x=c.x-b.x;temp2.y=c.y-b.y;temp2.z=c.z-b.z;
if(temp1.x*temp2.y==temp1.y*temp2.x&&temp1.x*temp2.z==temp1.z*temp2.x&&temp1.y*temp2.z==temp1.z*temp2.y)return 1;
return 0;
}
int check2(PO a,PO b,PO c,PO d)//判断共面;
{
int x1=b.x-a.x,y1=b.y-a.y,z1=b.z-a.z;
int x2=c.x-a.x,y2=c.y-a.y,z2=c.z-a.z;
int x3=d.x-a.x,y3=d.y-a.y,z3=d.z-a.z;
LL t1=(LL)x1*y2*z3+y1*z2*x3+z1*x2*y3;
LL t2=(LL)z1*x3*y2+x2*y1*z3+x1*z2*y3;
if(t1==t2)return 1;
return 0;
}
int dis(PO a,PO b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z);
}
int solve()
{
int sum1=0,sum2=0;
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
for(int k=1;k<=n;k++)
{
if(k==i||k==j)continue;
int dist=dis(po[k],po[i]);
if(dist!=dis(po[k],po[j]))continue;
if(check1(po[i],po[j],po[k]))continue;
for(int u=k+1;u<=n;u++)
{
if(u==i||u==j)continue;
if(dis(po[u],po[i])!=dis(po[u],po[j]))continue;
if(dis(po[u],po[i])!=dist)continue;
//if(i==1&&j==2&&k==3)cout<<check2(po[k],po[u],po[i],po[j])<<u<<"&%&^^*(\n";
if(check2(po[u],po[k],po[i],po[j]))continue;
if(dis(po[u],po[k])==dist&&dis(po[i],po[j])==dist)sum2++;
else sum1++;
}
}
}
}
return sum1/2+sum2/6;
}
int main()
{
int t,Case=0;
read(t);
while(t--)
{
read(n);
For(i,1,n)
{
read(po[i].x);read(po[i].y);read(po[i].z);
}
printf("Case #%d: %d\n",++Case,solve());
}
return 0;
}
hdu-5839 Special Tetrahedron(计算几何)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5776591.html