2016 CCPC网络选拔赛 部分题解

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
struct P {
    int x, y, z;
    P(int xx=0, int yy=0, int zz=0):x(xx), y(yy), z(zz) {}
    P operator-(const P& a) {
        return P(x-a.x, y-a.y, z-a.z);
    }
    P operator*(const P& b) {
        return P(y*b.z-z*b.y, z*b.x-x*b.z, x*b.y-y*b.x);
    }
    int dis() {
        return x*x + y*y + z*z;
    }
    void print() {
        printf("(%d %d %d)\n", x, y, z);
    }
}p[210];
int cross(const P& a, const P& b) {
    return a.x*b.x + a.y*b.y + a.z*b.z;
}

int main()
{
    int cas=0, t; cin>>t;
    while(cas<t) {
        int n;
        scanf("%d", &n);
        for(int i=0;i<n;i++) {
            scanf("%d %d %d", &p[i].x, &p[i].y, &p[i].z);
        }

        int ans = 0;
        for(int i=0;i<n;i++) {
            for(int j=i+1;j<n;j++) {
                for(int k=j+1;k<n;k++) {
                    int dis1 = (p[i]-p[j]).dis();
                    int dis2 = (p[i]-p[k]).dis();
                    int dis3 = (p[j]-p[k]).dis();
                    if(dis1!=dis2 && dis2!=dis3 && dis1!=dis3) continue;

                    int equa = -1; P t1, t2;
                    if(dis1==dis2 && dis1!=dis3) equa = dis1, t1 = p[j], t2 = p[k];
                    if(dis1==dis3 && dis1!=dis2) equa = dis1, t1 = p[i], t2 = p[k];
                    if(dis2==dis3 && dis1!=dis2) equa = dis2, t1 = p[i], t2 = p[j];

                    // if(equa==-1) equa = dis1;
                    for(int l=k+1;l<n;l++) {
                        if(cross((p[i]-p[j])*(p[i]-p[k]), p[i]-p[l])==0) continue;

                        if(equa!=-1) { 
                            if((p[l]-t1).dis()==equa && (p[l]-t2).dis()==equa) {
                                ++ans;
                            //    printf("%d %d %d %d\n", i, j, k, l);
                            }
                        } else { // 底面三边相等 
                            int dis4 = (p[l]-p[i]).dis();
                            int dis5 = (p[l]-p[j]).dis();
                            int dis6 = (p[l]-p[k]).dis();
                            if(dis4==dis1 && dis5==dis1 || dis4==dis1 && dis6==dis1 || dis5==dis1 && dis6==dis1) {
                                ++ans;
                            //    printf("%d %d %d %d\n", i, j, k, l);
                            }
                        }
                        
                    }
                }
            }
        }
        
        printf("Case #%d: %d\n", ++cas, ans);
    }
    return 0;
}