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
1
1
1 1
-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.
一道签到题,然而我却wa了很多次,我一直以为是让我判断长方形,发现还要输出面积。。。。
1 #include<iostream> 2 #include<cmath> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 struct node 7 { 8 int x,y; 9 }a[100]; 10 int main() 11 { 12 int n,i,j; 13 cin>>n; 14 for(i=1;i<=n;i++) 15 cin>>a[i].x>>a[i].y; 16 if(n==1) 17 { 18 cout<<"-1"<<endl; 19 } 20 else if(n==2) 21 { 22 if(a[1].x==a[2].x||a[1].y==a[2].y) 23 { 24 cout<<"-1"<<endl; 25 } 26 else 27 cout<<abs(a[1].x-a[2].x)*abs(a[2].y-a[1].y)<<endl; 28 } 29 else 30 { 31 for(i=1;i<=n;i++) 32 { 33 for(j=1;j<=n;j++) 34 { 35 if(a[i].x!=a[j].x&&a[i].y!=a[j].y) 36 { 37 cout<<abs(a[i].x-a[j].x)*abs(a[i].y-a[j].y)<<endl; 38 return 0; 39 } 40 } 41 } 42 cout<<"-1"<<endl; 43 } 44 return 0; 45 }