CodeForces - 596A Wilbur and Swimming Pool (模拟)

CodeForces - 596A

Wilbur and Swimming Pool

Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

Submit Status

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 x_i and y_i ( - 1000 ≤ x_i, y_i ≤ 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

20 01 1

Output

1

Input

11 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.

Source

Codeforces Round #331 (Div. 2)

//题意：

一个人要建一个矩形的游泳池，已经写好了矩形的四个顶点，他的朋友将他的几个顶点擦掉了，现在问剩下的几个点是否可以组成一个矩形，是的话输出矩形的面积，否则输出-1.

//思路：

就是一个模拟题，没啥技巧。。。

#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3f#define ll long long#define N 10010#define M 1000000007#define PI acos(-1.0)using namespace std;struct zz{int x;int y;}p[5];int cmp(zz a,zz b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;}int main(){int i,j,k;int n,m;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)scanf("%d%d",&p[i].x,&p[i].y);sort(p,p+n,cmp);if(n==1)printf("-1\n");else if(n==2){int y=abs(p[1].y-p[0].y);int x=abs(p[1].x-p[0].x);if(x==0||y==0)printf("-1\n");elseprintf("%d\n",x*y);}else if(n==3){int x=p[2].x-p[0].x;int y=max(abs(p[2].y-p[0].y),abs(p[1].y-p[0].y));if(x==0||y==0)printf("-1\n");elseprintf("%d\n",x*y);}else{int x=abs(p[3].x-p[0].x);int y=abs(p[3].y-p[0].y);if(x==0||y==0)printf("-1\n");elseprintf("%d\n",x*y);}}return 0;}