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  • uvaoj-1595:symmetry

    1595 - Symmetry

    The figure shown on the left is left-right symmetric as it is possible to fold the sheet of paper along a vertical line, drawn as a dashed line, and to cut the figure into two identical halves. The figure on the right is not left-right symmetric as it is impossible to find such a vertical line.

    epsfbox{p3226.eps}

    Write a program that determines whether a figure, drawn with dots, is left-right symmetric or not. The dots are all distinct.

    Input 

    The input consists of T test cases. The number of test cases T is given in the first line of the input file. The first line of each test case contains an integer N , where N ( 1$ le$N$ le$1, 000) is the number of dots in a figure. Each of the following N lines contains the x-coordinate and y-coordinate of a dot. Both x-coordinates and y-coordinates are integers between -10,000 and 10,000, both inclusive.

    Output 

    Print exactly one line for each test case. The line should contain `YES' if the figure is left-right symmetric. and `NO', otherwise.

    The following shows sample input and output for three test cases.

    Sample Input 

    3                                            
    5                                            
    -2 5                                         
    0 0 
    6 5 
    4 0 
    2 3 
    4 
    2 3 
    0 4 
    4 0 
    0 0 
    4 
    5 14 
    6 10
    5 10 
    6 14
    

    Sample Output 

    YES 
    NO 
    YES

    题解:刘汝佳白书135页练习题5.6;本题能够考虑暴力求解,也能够用set的查找来进行;


    code:

    #include <iostream>
    #include <set>
    #include <string>
    using namespace std;
    typedef pair<int,int> point;//定义类型;(不能够用预处理)

    set<point> se;

    int main()
    {
        int cas;
        int n;
        int x,y;
        cin>>cas;
        while(cas--)
        {
            se.clear();
            cin>>n;
            int sum=0;
            for(int i=0; i<n; i++)
            {
                cin>>x>>y;
                sum+=x;//将全部横坐标加和;
                se.insert(point(x*n,y));//(备注1)
            }
            /*
            set<point> ::iterator it;
            for(it=se.begin(); it!=se.end(); ++it)
            {
                point p=*it;
                cout<<p.first<<"  "<<p.second<<endl;
            }
            */
            set<point> ::iterator it;
            bool flag = true;
            for(it=se.begin(); it!=se.end(); ++it)
            {
                point p=*it;
                if(se.find(point(2*sum-p.first,p.second))==se.end())
                {
                    flag=false;
                    break;
                }
            }
            flag?

    cout<<"YES"<<endl: cout<<"NO"<<endl;
        }
        return 0;
    }


    备注1:

    x*n的意思,就是存储的时候将横坐标扩大n倍,由于对称轴肯定是竖线,所以如果对称的话全部的横坐标加和再与n想除得出的应该就是答案的x坐标;之所以不是将sum/n。是由于考虑到精度的问题。除的话会出现double类型;换而言之,这个存储方式就是将全部横坐标扩大了n倍;


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  • 原文地址:https://www.cnblogs.com/liguangsunls/p/7070227.html
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