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

A. Wilbur and Swimming Pool

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

2
0 0
1 1

output

1

input

1
1 1

output

-1

Note

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.

水题、但是窝过的也很水、想复杂了、题目意思是一个矩形抹去了若干点、前提是这本身就是个矩形，可是窝刚理解为任意四点、、、题意啊！写的还那么复杂、、渣

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;

struct node {
	int x, y;
}a[4];

int mianJi(int a1, int b1, int a2, int b2) {
	return abs(a1-a2) * abs(b1-b2);
}

int main() {
	int n;
	memset(a, 0, sizeof(a));

	cin >> n;
	for (int i = 0; i<n; i++)
		cin >> a[i].x >> a[i].y;
	if (n == 1)
		cout << -1 << endl;
	else {
		if (n == 2) {
			if (a[0].x != a[1].x && a[0].y != a[1].y)
				cout <<mianJi(a[0].x, a[0].y, a[1].x, a[1].y)<< endl;
			else
				cout << -1 << endl;
		}
		else if (n == 3) {
			int flag1 = 0;
			int flag2 = 0;
			for (int i = 0; i<3; i++) {
				for (int j = i+1; j<3; j++) {
					if (a[i].x == a[j].x)
						flag1 = 1;
					if (a[i].y == a[j].y)
						flag2 = 1;
				}
			}
			if (flag1 && flag2) {
				for (int i = 0; i<3; i++) {
					for (int j = i+1; j<3; j++) {
						if (a[i].x != a[j].x && a[i].y != a[j].y)
							cout << mianJi(a[i].x, a[i].y, a[j].x, a[j].y) << endl;
					}
				}
			}
			else
				cout << -1 << endl;
		}
		else {
			int flag3 = 0;
			int flag4 = 0;
			for (int i = 0; i<4; i++) {
				for (int j = i+1; j<4; j++) {
					if (a[i].x == a[j].x)
						flag3 ++ ;
					if (a[i].y == a[j].y)
						flag4 ++ ;
				}
			}
			if (flag3 == 2 && flag4 == 2) {
				for (int i = 0; i<3; i++) {
					for (int j = i+1; j<3; j++) {
						if (a[i].x != a[j].x && a[i].y != a[j].y)
							cout << mianJi(a[i].x, a[i].y, a[j].x, a[j].y) << endl;
					}
				}
			}
			else
				cout << -1 << endl;
		}
	}
	return 0;
}

一下附上经典代码吧、

#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==1)return puts("-1");
    for(int i=1;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));
}

多想多思考、看到题目不能脑海里浮现出思路就开始码代码，还要思考一下自己的想法是最优的或者可行与否、是否还存在更优的求解方案！？？这样才会避免过多的wa！都这么长时间了还停留在div2的AB上也的确要反洗自己了、、、、