C++ 检查四个线段是否形成一个矩形(Check if four segments form a rectangle)-优快云博客

本文链接：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/145686317

我们有四条线段作为其端点的一对坐标。我们需要判断这四条线段是否构成矩形。
示例：

输入：segment[] = [(4, 2), (7, 5),
(2, 4), (4, 2),
(2, 4), (5, 7),
(5, 7), (7, 5)]
输出：
是给定这些线段，可以组成一个长度为 3X2 的矩形。

输入：segment[] = [(7, 0), (10, 0),
(7, 0), (7, 3),
(7, 3), (10, 2),
(10, 2), (10, 0)]
输出：否
这些线段不能组成矩形。

上述示例如下所示：
这个问题主要是：如何检查给定的四个点是否形成一个正方形
JavaScript：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/145686032
C#：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/145686008
Python：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/145685981
Java：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/145685948
C++：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/145685863

我们可以通过使用矩形的属性来解决这个问题。首先，我们检查线段的唯一端点总数，如果这些点的数量不等于 4，则线段不能构成矩形。然后我们检查所有点对之间的距离，最多应该有 3 个不同的距离，一个用于对角线，两个用于边，最后我们将检查这三个距离之间的关系，对于构成矩形的线段，这些距离应该满足勾股关系，因为矩形的边和对角线构成直角三角形。如果它们满足上述条件，那么我们将线段构成的多边形标记为矩形，否则不是。

示例代码：

// C++ program to check whether it is possible
// to make a rectangle from 4 segments
#include <bits/stdc++.h>
using namespace std;
#define N 4

// structure to represent a segment
struct Segment
{
int ax, ay;
int bx, by;
};

// Utility method to return square of distance
// between two points
int getDis(pair<int, int> a, pair<int, int> b)
{
return (a.first - b.first)*(a.first - b.first) +
(a.second - b.second)*(a.second - b.second);
}

// method returns true if line Segments make
// a rectangle
bool isPossibleRectangle(Segment segments[])
{
set< pair<int, int> > st;

// putting all end points in a set to
// count total unique points
for (int i = 0; i < N; i++)
{
st.insert(make_pair(segments[i].ax, segments[i].ay));
st.insert(make_pair(segments[i].bx, segments[i].by));
}

// If total unique points are not 4, then
// they can't make a rectangle
if (st.size() != 4)
return false;

// dist will store unique 'square of distances'
set<int> dist;

// calculating distance between all pair of
// end points of line segments
for (auto it1=st.begin(); it1!=st.end(); it1++)
for (auto it2=st.begin(); it2!=st.end(); it2++)
if (*it1 != *it2)
dist.insert(getDis(*it1, *it2));

// if total unique distance are more than 3,
// then line segment can't make a rectangle
if (dist.size() > 3)
return false;

// copying distance into array. Note that set maintains
// sorted order.
int distance[3];
int i = 0;
for (auto it = dist.begin(); it != dist.end(); it++)
distance[i++] = *it;

// If line seqments form a square
if (dist.size() == 2)
return (2*distance[0] == distance[1]);

// distance of sides should satisfy pythagorean
// theorem
return (distance[0] + distance[1] == distance[2]);
}

// Driver code to test above methods
int main()
{
Segment segments[] =
{
{4, 2, 7, 5},
{2, 4, 4, 2},
{2, 4, 5, 7},
{5, 7, 7, 5}
};

(isPossibleRectangle(segments))?cout << "Yes\n":cout << "No\n";
}