Codeforces Round #331 (Div. 2) A. Wilbur and Swimming Pool 水题

时间：2022-06-11 11:34:19

A. Wilbur and Swimming Pool

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/596/problem/A

Description

After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the area of the rectangle must be positive. Wilbur had all four vertices of the planned pool written on a paper, until his friend came along and erased some of the vertices.

Now Wilbur is wondering, if the remaining n vertices of the initial rectangle give enough information to restore the area of the planned swimming pool.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 4) — the number of vertices that were not erased by Wilbur's friend.

Each of the following n lines contains two integers xi and yi ( - 1000 ≤ xi, yi ≤ 1000) —the coordinates of the i-th vertex that remains. Vertices are given in an arbitrary order.

It's guaranteed that these points are distinct vertices of some rectangle, that has positive area and which sides are parallel to the coordinate axes.

Output

Print the area of the initial rectangle if it could be uniquely determined by the points remaining. Otherwise, print  - 1.

Sample Input

2
0 0
1 1

Sample Output

1

HINT

题意

一个矩形，4个点，但是被人擦去了，现在就只有n个点了

然后问你这n个点能不能构成独一无二的矩形，如果可以输出面积

否则否则输出-1

题解：

扫一遍四个点，判断minx==maxx miny==maxy，满足两个中的一个，就输出-1

否则输出面积就好了

代码

#include<iostream>

#include<stdio.h>

#include<algorithm>

using namespace std;

int main()

{

    int n;cin>>n;

    int minx,maxx,miny,maxy;

    cin>>minx>>miny;

    maxx=minx,maxy=miny;

    if(n==)return puts("-1");

    for(int i=;i<n;i++)

    {

        int x,y;cin>>x>>y;

        minx=min(minx,x);

        maxx=max(maxx,x);

        miny=min(miny,y);

        maxy=max(maxy,y);

    }

    if((maxx==minx)||(maxy==miny))

        return puts("-1");

    printf("%d\n",(maxx-minx)*(maxy-miny));

}

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