Fzu2270 Two Triangles

题意:n个点(n<=10),取出6个点,组成两个一样的三角形,一样的三角形定义为旋转,平移后一样

题解:直接枚举6个点,6个点取三角形,先判断是不是三角形,再判断点的顺序和边的大小

#include<stdio.h>
#include<bitset>
#include<iostream>
#include<algorithm>
#include<vector>
#include<string.h>
#define INF 0x3f3f3f3f
#define maxn 15
using namespace std;
struct node{
    int x, y;
    node(){x=y=0;}
    node(int aa,int bb){x = aa,y = bb;}
    node(const node& rhx){x = rhx.x;y = rhx.y;}
    friend node operator - (const node& aa,const node& bb){return node(aa.x-bb.x, aa.y-bb.y);}
    friend node operator + (const node& aa,const node& bb){return node(aa.x+bb.x, aa.y+bb.y);}
}a[maxn], c1[maxn], c2[maxn];
int t1[maxn], t2[maxn], ans, T, n, b[maxn], ca = 1;
int dis(node &aa,node &bb){
    return (aa.x-bb.x)*(aa.x-bb.x)+(aa.y-bb.y)*(aa.y-bb.y);
}
int cross(const node &aa,const node &bb){
    return aa.x*bb.y-aa.y*bb.x;
}
int cmp(node &aa,node &bb){
    if(aa.x == bb.x) return aa.y>bb.y;
    return aa.x<bb.x;
}
double area(node *c) {
    int area = 0;
    area += cross(c[1]-c[0], c[2]-c[0]);
    return area;
}
int check(){
    if(cross(c1[0]-c1[1], c1[0]-c1[2]) == 0||cross(c2[0]-c2[1], c2[0]-c2[2]) == 0) return 0;
    if(area(c1)*area(c2) < 0){
        swap(c1[0], c1[2]);
    }
    for(int i=0;i<3;i++){
        t1[i] = dis(c1[i], c1[(i+1)%3]);
        t2[i] = dis(c2[i], c2[(i+1)%3]);
    }
    if((t1[0] == t2[0]&&t1[1] == t2[1]&&t1[2] == t2[2])
    ||(t1[0] == t2[1]&&t1[1] == t2[2]&&t1[2] == t2[0])
    ||(t1[0] == t2[2]&&t1[1] == t2[0]&&t1[2] == t2[1]))
        return 1;
    return 0;
}
int main(){
    cin>>T;
    while(T--){
        ans = 0;
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>a[i].x>>a[i].y;
        int comb = (1<<6)-1, x, y;
        while(comb < (1<<n)){
            int num = 0;
            for(int i=0;i<n;i++)
                if(comb&(1<<i)) b[num++] = i;
            int t = (1<<3)-1;
            while(t < (1<<6)){
                int num1=0,num2=0;
                for(int i=0;i<6;i++)
                  if(t&(1<<i)) c1[num1++] = a[b[i]];
                  else c2[num2++] = a[b[i]];
                if(check())
                    ans++;
                x = t&-t;y = t+x;
                t = ((t&~y)/x>>1)|y;
            }
            x = comb&-comb;y = comb+x;
            comb = ((comb&~y)/x>>1)|y;
        }
        printf("Case %d: ", ca++);
        cout<<ans<<endl;
    }
    return 0;
}