#1040 : 矩形判断

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#1040 : 矩形判断

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

给出平面上4条线段，判断这4条线段是否恰好围成一个面积大于0的矩形。

输入

输入第一行是一个整数T(1<=T<=100)，代表测试数据的数量。

每组数据包含4行，每行包含4个整数x1, y1, x2, y2 (0 <= x1, y1, x2, y2 <= 100000)；其中(x1, y1), (x2,y2)代表一条线段的两个端点。

输出

每组数据输出一行YES或者NO，表示输入的4条线段是否恰好围成矩形。

样例输入

30 0 0 11 0 1 10 1 1 11 0 0 00 1 2 31 0 3 23 2 2 31 0 0 10 1 1 01 0 2 02 0 1 11 1 0 1

样例输出

YESYESNO

#include <iostream>#include <limits.h>#include <set>using namespace std;struct Line {int x1, y1, x2, y2;};int main(){int T;cin >> T;set<pair<int, int>> st;while (T--) {st.clear();Line line1, line2, line3, line4;cin >> line1.x1 >> line1.y1 >> line1.x2 >> line1.y2;st.insert(make_pair(line1.x1, line1.y1));st.insert(make_pair(line1.x2, line1.y2));cin >> line2.x1 >> line2.y1 >> line2.x2 >> line2.y2;st.insert(make_pair(line2.x1, line2.y1));st.insert(make_pair(line2.x2, line2.y2));cin >> line3.x1 >> line3.y1 >> line3.x2 >> line3.y2;st.insert(make_pair(line3.x1, line3.y1));st.insert(make_pair(line3.x2, line3.y2));cin >> line4.x1 >> line4.y1 >> line4.x2 >> line4.y2;st.insert(make_pair(line4.x1, line4.y1));st.insert(make_pair(line4.x2, line4.y2));if (st.size() != 4) {cout << "NO" << endl;continue;}double slope1, slope2, slope3, slope4;slope1 = (line1.x1 == line1.x2) ? INT_MAX : (double)(line1.y2 - line1.y1) / (line1.x2 - line1.x1);slope2 = (line2.x1 == line2.x2) ? INT_MAX : (double)(line2.y2 - line2.y1) / (line2.x2 - line2.x1);slope3 = (line3.x1 == line3.x2) ? INT_MAX : (double)(line3.y2 - line3.y1) / (line3.x2 - line3.x1);slope4 = (line4.x1 == line4.x2) ? INT_MAX : (double)(line4.y2 - line4.y1) / (line4.x2 - line4.x1);if (slope1 == slope2) {if (slope3 != slope4) {cout << "NO" << endl;continue;}//垂直if (((line1.y1 - line1.y2) * (line3.y1 - line3.y2) +(line1.x1 - line1.x2) * (line3.x1 - line3.x2)) == 0) {cout << "YES" << endl;}else {cout << "NO" << endl;}}///////////////////////////////////else if (slope1 == slope3) {if (slope2 != slope4) {cout << "NO" << endl;continue;}//垂直if (((line1.y1 - line1.y2) * (line2.y1 - line2.y2) +(line1.x1 - line1.x2) * (line2.x1 - line2.x2)) == 0) {cout << "YES" << endl;}else {cout << "NO" << endl;}}/////////////////////////////////else if(slope1 == slope4) {if (slope2 != slope3) {cout << "NO" << endl;continue;}//垂直if (((line1.y1 - line1.y2) * (line3.y1 - line3.y2) +(line1.x1 - line1.x2) * (line3.x1 - line3.x2)) == 0) {cout << "YES" << endl;}else {cout << "NO" << endl;}}/////////////////////////////else {cout << "NO" << endl;}}return 0;}

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