Run

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 101    Accepted Submission(s): 43

Problem Description

AFA is a girl who like runing.Today,he download an app about runing .The app can record the trace of her runing.AFA will start runing in the park.There are many chairs in the park,and AFA will start his runing in a chair and end in this chair.Between two chairs,she running in a line.she want the the trace can be a regular triangle or a square or a regular pentagon or a regular hexagon.
Please tell her how many ways can her find.
Two ways are same if the set of chair that they contains are same.

Input

There are multiply case.
In each case,there is a integer n(1 < = n < = 20)in a line.
In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.

Output

Output the number of ways.

Sample Input

4
0 0
0 1
1 0
1 1

Sample Output

1

虽然题目很简单，我还是来个通用的代码吧

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <bitset>
#include <sstream>
#include <cstdlib>
#include <complex>
#include <climits>
#include <cstring>
#include <iostream>
#include <algorithm>

#define REP(i,N) for (int i = 0;i < (N);i++)
#define REP_1(i,N) for (int i = 1;i < (N);i++)
#define REP_2(i,be,en) for (int i = (be);i < (en);i++)
#define DWN(i,N) for (int i = (N);i >= 0;i--)
#define DWN_1(i,N) for (int i = (N);i >= 1;i--)
#define DWN_2(i,en,be) for (int i = (en);i >= (be);i--)
#define FR(N) freopen((N),"r",stdin)
#define FW(N) freopen((N),"w",stdout)
#define SO(N) scanf("%d",&(N))
#define STO(N,M) scanf("%d %d",&(N),&(M))
#define STH(N,M,K) scanf("%d %d %d",&(N),&(M),&(K))
#define PS(N) printf("%d",(int)(N));
#define MAXN 2010
#define GETS(ch) fgets((ch),MAXN,stdin)
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;

typedef long long LL;
typedef LL ll;
typedef map<int,LL> MINT;
typedef map<char,int> MCH;
typedef map<string,int> MSTR;

typedef vector<int> VINT;
typedef set<LL> SINT;
typedef pair<int,int> PINT;

LL read(){
    LL ret=0,f=1;
    char x=getchar();
    while(!(x>=‘0‘ && x<=‘9‘)){if(x==‘-‘)f=-1;x=getchar();}
    while(x>=‘0‘ && x<=‘9‘) ret=ret*10+x-‘0‘, x=getchar();
    return ret*f;
}

class point {
public:
    int x,y;
}node[30];
point List[5];

int cross(point p0,point p1,point p2) //计算叉积  p0p1 X p0p2
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
double dis(point p1,point p2)  //计算 p1p2的 距离
{
    return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
}
bool cmp(point p1,point p2) //极角排序函数 ， 角度相同则距离小的在前面
{
    int tmp=cross(List[0],p1,p2);
    if(tmp>0) return true;
    else if(tmp==0&&dis(List[0],p1)<dis(List[0],p2)) return true;
    else return false;
}

int cross2(point p0,point p1,point p2) //计算点积  p0p1 · p1p2
{
    return (p1.x-p0.x)*(p2.x-p1.x)+(p1.y-p0.y)*(p2.y-p1.y);
}

int ans = 0;
int work(vector<int> V) {
    int len = V.size();
    for (int i = 1;i < len;i++) {
        if (node[V[i]].x < node[V[0]].x) {
            swap(V[i],V[0]);
        }
        else if (node[V[i]].x == node[V[0]].x) {
            if (node[V[i]].y < node[V[0]].y) {
                swap(V[i],V[0]);
            }
        }
    }
    for (int i = 0;i < len;i++) {
        List[i] = node[V[i]];
    }
    sort(List,List + len,cmp);
    for (int i = 0;i < len - 1;i++) {
        if (dis(List[i],List[i + 1]) - dis(List[i + 1],List[(i + 2) % len]) > 1e-6) return 0;
        if (cross2(List[(i + len - 1) % len],List[i],List[i + 1]) != 0) return 0;
    }

    return 1;
}
int N;

void dfs(int u,int pos,vector<int> V,int num) {
    if (pos == num) {
        ans += work(V);
        return;
    }
    if (u >= N) return;
    V.push_back(u);
    dfs(u + 1,pos + 1,V,num);
    V.pop_back();
    dfs(u + 1,pos,V,num);
}

int main() {
    while (cin >> N) {
        for (int i = 0;i < N;i++) {
            cin >> node[i].x >> node[i].y;
        }
        vector<int> V;
        V.clear();
        ans = 0;
        dfs(0,0,V,4); // 只能形成4边形
        cout << ans << endl;
    }
}