zoukankan      html  css  js  c++  java
  • 【CodeForces 596A】E

    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
    ",area);
        else printf("-1
    ");
        return 0;
    }

      

  • 相关阅读:
    2w字 + 40张图带你参透并发编程!
    完了,这个硬件成精了,它竟然绕过了 CPU...
    一文详解 Java 并发模型
    详解匈牙利算法与二分图匹配
    机器学习 | 详解GBDT在分类场景中的应用原理与公式推导
    Python | 浅谈并发锁与死锁问题
    LeetCode 91,点赞和反对五五开,这题是好是坏由你来评判
    LeetCode 90 | 经典递归问题,求出所有不重复的子集II
    【Azure DevOps系列】什么是Azure DevOps
    MSIL入门(四)之委托delegate
  • 原文地址:https://www.cnblogs.com/flipped/p/5045326.html
Copyright © 2011-2022 走看看