Finding the Order

链接：https://ac.nowcoder.com/acm/contest/5669/F
来源：牛客网

 

题目描述

ZYB has a so-called smart brain: He can always point out the keypoint in a complex problem.
There are two parallel lines AB and CD in a plane. A,B,C,D{A,B,C,D}A,B,C,D are all distinct points.  You only know the Euclidean Distances between AC,AD,BC,BD{AC,AD,BC,BD}AC,AD,BC,BD. but you don’t know the exact order of points. (i.e. You don’t know whether it’s AB∥CDAB \parallel CDAB∥CD or AB∥DCAB \parallel DCAB∥DC).

Could you determine the order of points quickly, like the ZYB does?

输入描述:

The input contains multiple cases. The first line of the input contains a single integer T (1≤T≤100)T\ (1 \le T \le 100)T (1≤T≤100), the number of cases.
For each case, there are four integers a,b,c,d(1≤a,b,c,d≤1000)a,b,c,d(1 \le a,b,c,d \le 1000)a,b,c,d(1≤a,b,c,d≤1000) in a line, indicating the distances between AC,AD,BC,BD{AC,AD,BC,BD}AC,AD,BC,BD.It is guaranteed that each case corresponds to a valid solution.

输出描述:

For each case, output 'AB//CD' (Quotation marks) if AB∥CDAB \parallel CDAB∥CD, or output 'AB//DC' (Quotation marks) if AB∥DCAB \parallel DCAB∥DC.

示例1

输入

复制 2 3 5 5 3 5 3 3 5

2
3 5 5 3
5 3 3 5

输出

复制 AB//CD AB//DC

AB//CD
AB//DC

---

假设C在D前面，现在把D慢慢移动到C左边，这样发现对角线和边界线交叉了，往对角线和边界线上去想

标出对角线的点，两个对角线相加是对角点左右两个三角形的两边长度和，满足三角形两边之和大于第三边，这就是C在D前面的特征

这个题的启发是固定当前规格，动态地移动点去找规律。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <math.h>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define ll long long
#define Inf 0x3f3f3f3f
using namespace std;
const int maxn=1e5+5;
int a,b,c,d;
int main(){
    int T;
    cin>>T;
    while(T--)
    {     
        cin>>a>>b>>c>>d;
        if(b+c>a+d){
            cout<<"AB//CD\n";
        }else{
            cout<<"AB//DC\n";
        }
    }
    return 0;
}