专栏首页饶文津的专栏【CodeForces 596A】E - 特别水的题5-Wilbur and Swimming Pool

【CodeForces 596A】E - 特别水的题5-Wilbur and Swimming Pool

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

Input

2 0 0 1 1

Output

1

Input

1 1 1

Output

-1

Hint

In the first sample, two opposite corners of the initial rectangle are given, and that gives enough information to say that the rectangle is actually a unit square.

In the second sample there is only one vertex left and this is definitely not enough to uniquely define the area.

这题可以讨论n为1、2、3、4的情况,n=3或者4时用循环找出x,y坐标不同的两点,面积要绝对值。

或者不要讨论,直接在读入的时候,找出x,y的最大值和最小值,计算面积如果面积==0,,那就输出-1,代码如下:

#include<stdio.h>
#include<algorithm>
using namespace std;
long long n,area,xmin=1001,ymin=1001,xmax=-1001,ymax=-1001;
struct Node
{
    long long x,y;
} node[5];
int main()
{
    scanf("%lld",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%lld%lld",&node[i].x,&node[i].y);
        xmax=max(xmax,node[i].x);
        xmin=min(xmin,node[i].x);
        ymax=max(ymax,node[i].y);
        ymin=min(ymin,node[i].y);
    }
    area=(ymax-ymin)*(xmax-xmin);
    if(area)printf("%lld\n",area);
    else printf("-1\n");
    return 0;
}

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