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.

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 ( 1N1,
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；