Greeting Card

最新推荐文章于 2019-04-15 20:26:25 发布

原创最新推荐文章于 2019-04-15 20:26:25 发布 · 352 阅读

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本文探讨了一个复杂的绘图挑战，旨在通过精确计算点之间的距离来确定能够形成特定长度段的点对数量。利用整数坐标的特性，文章提出了一种高效的算法解决方案，避免了超时问题，确保了足够的墨水供应以完成新年贺卡的打印任务。

题目描述

Quido plans to send a New Year greeting to his friend Hugo. He has recently acquired access to an advanced high-precision plotter and he is planning to print the greeting card on the plotter.
Here’s how the plotter operates. In step one, the plotter plots an intricate pattern of n dots on the paper. In step two, the picture in the greeting emerges when the plotter connects by a straight segment each pair of dots that are exactly 2 018 length units apart.
The plotter uses a special holographic ink, which has a limited supply.Quido wants to know the number of all plotted segments in the picture to be sure that there is enough ink to complete the job.

输入

The first line of input contains a positive integer n specifying the number of plotted points. The following n lines each contain a pair of space-separated integer coordinates indicating one plotted point. Each coordinate is non-negative and less than 231. There are at most 105 points, all of them are distinct.
In this problem, all coordinates and distances are expressed in plotter length units, the length of the unit in the x-direction and in the y-direction is the same.

输出

The output contains a single integer equal to the number of pairs of points which are exactly 2018 length units apart.

题意：

给出一组点，求其中两点之间距离为2018的数量。

题解：

刚开始用暴力写但数据太大死在了超时上，因为点的坐标都为整数打表可以找到规律。只存在两种情况1.在同一直线上相距2018，二、分别相差1118和1680。用其中一个点根据关系遍历看该点在集合中是否存在，即可以找到所有满足的点。

#include<iostream>
#include<cstdio>
#include<set>
#include<utility>
typedef long long ll;

using namespace std;
const int maxn = 100000+2;
int n;
ll x,y;
ll ans;
int dx[]={2018,0,-2018,0,1118,1680,1680,-1118,-1680,1118,-1118,-1680};
int dy[]={0,2018,0,-2018,1680,1118,-1118,1680,1118,-1680,-1680,-1118};
ll X[maxn],Y[maxn];
int main()
{

    cin>>n;
    pair<ll,ll> p;
    set<pair<ll,ll> >st;
    for(int i = 0;i<n;i++){
        cin>>X[i]>>Y[i];
        p.first = X[i];
        p.second = Y[i];
        st.insert(p);
    }
    ans = 0;
    for(int i = 0;i<n;i++){
        for(int j = 0;j<12;j++){
            p.first = X[i]+dx[j];
            p.second = Y[i]+dy[j];
            if(st.count(p))
                ans++;
        }
    }
    cout<<ans/2<<endl;
    return 0;

}