zoukankan      html  css  js  c++  java
  • Parallelogram Counting(平行四边形个数,思维转化)

    1058 - Parallelogram Counting
    Time Limit: 2 second(s) Memory Limit: 32 MB

    There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points that can be written as {A, B, C, D} such that AB || CD, and BC || AD. No four points are in a straight line.

    Input

    Input starts with an integer T (≤ 15), denoting the number of test cases.

    The first line of each test case contains an integer n (1 ≤ n ≤ 1000). Each of the next n lines, contains 2 space-separated integersx and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000.

    Output

    For each case, print the case number and the number of parallelograms that can be formed.

    Sample Input

    Output for Sample Input

    2

    6

    0 0

    2 0

    4 0

    1 1

    3 1

    5 1

    7

    -2 -1

    8 9

    5 7

    1 1

    4 8

    2 0

    9 8

    Case 1: 5

    Case 2: 6

    题解:

    给一系列点,让找到可以组成的平行四边形的个数;

    思路:由于平行四边形的交点是唯一的,那么我们只要记录每两个直线的交点,判断交点重复的个数可以求出来了;假设有这个交点有三个重复的,则ans+=3*2/2;

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    const int INF=0x3f3f3f3f;
    typedef long long LL;
    const int MAXN=1010;
    struct Dot{
        double x,y;
        bool operator < (const Dot &b) const{
            if(x!=b.x)return x<b.x;
            else return y<b.y;
        }
    };
    Dot dt[MAXN];
    Dot num[MAXN*MAXN];
    int main(){
        int T,N,kase=0;
        scanf("%d",&T);
        while(T--){
            scanf("%d",&N);
            for(int i=0;i<N;i++)scanf("%lf%lf",&dt[i].x,&dt[i].y);
            int tp=0;
            for(int i=0;i<N;i++)
                for(int j=i+1;j<N;j++){
                    if(dt[i].x==dt[j].x&&dt[i].y==dt[j].y)continue;
                    num[tp].y=(dt[i].y+dt[j].y)/2;
                    num[tp].x=(dt[i].x+dt[j].x)/2;
                    tp++;
                }
                sort(num,num+tp);
                LL ans=0,cur=1;
            for(int i=1;i<tp;i++){
                if(num[i].x==num[i-1].x&&num[i].y==num[i-1].y)cur++;
                else{
                    ans+=cur*(cur-1)/2;cur=1;
                }
            }
            printf("Case %d: %lld
    ",++kase,ans);
        }
        return 0;
    }
  • 相关阅读:
    【Swift】图文混排,ios开发中在textfield或textView中插入图片
    ios开发-指纹识别
    ios开发-程序压后台后,悄悄的抓取数据~~
    setNeedDisplay和setNeedsLayout
    rangeOfString用法
    NSThread的使用
    UIActivityIndicatorView的详细使用
    iOS高斯模糊处理
    HIbernate学习笔记(一) 了解hibernate并搭建环境建立第一个hello world程序
    Hibernate4搭建Log4J日志管理(附Log4j.properties配置详解)
  • 原文地址:https://www.cnblogs.com/handsomecui/p/5259374.html
Copyright © 2011-2022 走看看