light oj 1058

标签：

哈希老坑了。。。

问你用已知点能找出多少个平行四边形。

原来想着哈希，哪知道数据很小，哈希的话老是不对，一看大牛们都是直接排序。。。瞎了。。。慎用哈希。。。慎用浮点。。。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
const int MAXN=1010;
const int INF = (1 << 30) - 1;
const double eps = 1e-10;
double add(double a,double b) //有误差的浮点加法
{
    if(abs(a + b) < eps * (abs(a) + abs(b))) return 0;
    return a + b;
}
struct Point
{
    int x,y;
    Point(double tx = 0,double ty = 0) : x(tx),y(ty) {}
    Point operator + (Point p)
    {
        return Point(add(x,p.x),add(y,p.y));
    }
    Point operator - (Point p)
    {
        return Point(add(x,-p.x),add(y,-p.y));
    }
    Point operator * (double d)
    {
        return Point(x * d,y * d);
    }
    Point operator / (double d)
    {
        return Point(x / d,y / d);
    }
    void Read()
    {
        scanf("%d%d",&x,&y);
    }
};
Point date[MAXN*1000],lis[MAXN*1000];
int Sign(double x)
{
    return (x > eps) - (x < -eps);
}

double Cross(Point a,Point b,Point c)
{
    return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}

double Dis(Point a,Point b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
bool cmp(Point a,Point b)
{
    if (a.x==b.x) return a.y<b.y;
    else return a.x<b.x;
}
int main()
{
    int T,cas=1;
    scanf ("%d",&T);
    while (T--)
    {
        int n;
        scanf ("%d",&n);
        for (int i=0; i<n; i++)
            date[i].Read();
        int top=0;
        for (int i=0; i<n; i++)
        {
            for (int j=i+1; j<n; j++)
            {
                lis[top].x=date[i].x+date[j].x;
                lis[top].y=date[i].y+date[j].y;
                top++;
            }
        }
        sort(lis,lis+top,cmp);
        int ans=0,temp=1;
        for (int i=1; i<top; i++)
        {
            if (lis[i].x==lis[i-1].x&&lis[i].y==lis[i-1].y)
            {
                temp++;
            }
            else
            {
                
                temp=1;
            }
            ans+=temp-1;
        }
        printf ("Case %d: %d\n",cas++,ans);
    }
    return 0;
}

light oj 1058

标签：

原文地址：http://www.cnblogs.com/nj-czy/p/5286784.html