Problem 2270 Two Triangles
Accept: 73 Submit: 230
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Fat brother and Maze are playing a kind of special (hentai) game with some points in a two-dimensional plane. First Fat Brother would choose three different points and then Maze would choose other three different points. If both of the player’s points form a triangle, they would feel happy. Also, if these two triangles are exactly the same, they would feel even happier. The two triangles are considered the same if one can be changed from another by translation and rotation.
Now given N points in a plane, Fat Brother would like to know how many different sets of points they can choose in order to make them even happier.
Input
The first line of the data is an integer T (1 <= T <= 100), which is the number of the text cases.
Then T cases follow, each case contains an integer N (6 <= N <= 10) indicated the number of the points in the two-dimensional plane. The N lines follow, each line contains two integers x and y (0 <= x, y <= 10) shows this point’s coordinate. All the points have different coordinate in this plane.
Output
For each case, output the case number first, and then output the number of the different sets of points they can choose in order to make them even happier.
Sample Input
Sample Output
Hint
In the first case, Fat Brother can choose a set which contains first three points and Maze can choose a set which contains last three points, this makes a suitable answer. Also, Fat Brother and Maze can change the set they choose and still make a solid result.
Source
第七届福建省大学生程序设计竞赛-重现赛（感谢承办方闽江学院）
所以首先要是三角形即三点不能共线，其次要全等，即三条边对应相等，最后判断是否对称，如果对称无论如何旋转都无法重合
所以利用叉积选好对应边，如果两个叉积互为相反数，则说明对称否则可以重合，这里注意一种特殊情况，等边三角形要提前判断，因为等边三角形不需要判断叉积只要全等一定可以通过旋转平移重合
直接暴力枚举六个不同的点即可
代码：
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstring>
#include<iostream>
#include<sstream>
#include<cmath>
#define LL long long
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
int n;
int T;
int ans;
struct node
{
int x,y;
node (){};
node(int x,int y):x(x),y(y){};
node operator - (node p)
{
return node(x-p.x,y-p.y);
}
int operator *(node p){
return x*p.y-y*p.x;
}
}E[15];
int vis[11][11][11][11][11][11];
double dis(node a,node b)
{
return sqrt((a.x-b.x)*(a.x-b.x)*1.0+(a.y-b.y)*(a.y-b.y)*1.0);
}
int check(int a,int b,int c,int d,int e,int f){
int a1[3] = {a,b,c};
int a2[3] = {d,e,f};
sort(a1,a1+3);
sort(a2,a2+3);
if(vis[a1[0]][a1[1]][a1[2]][a2[0]][a2[1]][a2[2]]==0){
vis[a1[0]][a1[1]][a1[2]][a2[0]][a2[1]][a2[2]]=1;
return 1;
}
else
return 0;
}
int dis2(node a,node b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int main()
{
scanf("%d",&T);
int kase = 1;
while(T--){
ans = 0;
scanf("%d",&n);
for(int i = 0;i<n;i++){
scanf("%d%d",&E[i].x,&E[i].y);
}
memset(vis,0,sizeof(vis));
for(int i = 0;i<n;i++){
for(int j = 0;j<n;j++){
if(i==j)
continue;
for(int k = 0;k<n;k++){
if(i==k||j==k)
continue;
for(int m = 0;m<n;m++){
if(fabs(dis(E[i],E[k])-dis(E[j],E[m]))>eps)
continue;
if(m==i||m==j||m==k)
continue;
for(int l = 0;l<n;l++)
{
if(l==i||l==j||l==k||l==m)
continue;
for(int o = 0;o<n;o++){
if(fabs(dis(E[i],E[l])-dis(E[j],E[o]))>eps||fabs(dis(E[k],E[l])-dis(E[m],E[o]))>eps)
continue;
if(o==i||o==j||o==k||o==m||o==l)
continue;
if((((E[i]-E[k])*(E[i]-E[l]))*((E[j]-E[m])*(E[j]-E[o])))<0)
continue;
double len1[3];
double len2[3];
len1[0] = dis(E[i],E[k]);
len1[1] = dis(E[i],E[l]);
len1[2] = dis(E[k],E[l]);
len2[0] = dis(E[j],E[m]);
len2[1] = dis(E[j],E[o]);
len2[2] = dis(E[m],E[o]);
sort(len1,len1+3);
sort(len2,len2+3);
if(fabs(len1[2]-len1[0]-len1[1])<=eps)
continue;
if(fabs(len2[2]-len2[0]-len2[1])<=eps)
continue;
// if(!check(i,k,l,j,m,o))
// continue;
ans++;
}
}
}
}
}
}
printf("Case %d: %d\n",kase++,ans/6);
}
}