UVA 1595对称轴 set集合

最新推荐文章于 2021-04-12 19:04:07 发布

hanker99

最新推荐文章于 2021-04-12 19:04:07 发布

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本文介绍了一种用于判断由点绘制的图形是否左右对称的程序。通过输入一系列点坐标，程序能够确定图形沿垂直轴折叠后是否能形成两个完全相同的部分。此算法适用于计算机图形学和几何形状分析。

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 (1 ≤ N ≤ 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.
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

#include<bits/stdc++.h>
using namespace std;

struct Point{
    double x,y;
    Point(double a,double b){ x=a,y=b;}
    bool operator < (const Point & b)const{
        if(x!=b.x){
            return x<b.x;
        }
        else return  y<b.y;
    }
    bool operator == (const Point & b)const{
        return (x==b.x&&y==b.y);
    }
};

typedef pair<double,double> Pair;

set<Point> s;

int main(){
    //freopen("datain.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--){
        s.clear();
        int n;
        scanf("%d",&n);
        double xSum=0;
        for(int i=0;i<n;i++){
            double x,y;
            scanf("%lf%lf",&x,&y);
            Point a (x,y);
            s.insert(a);
            xSum+=x;
        }
        double axis_x=xSum/n;
        bool good=true;
        for(set<Point>::iterator it=s.begin();it!=s.end();++it){
            if(it->x==axis_x) continue;
            else if(it->x < axis_x){
                double right=(axis_x - it->x)+axis_x;
                Point b(right,it->y);
                if(!s.count(b)) {
                    good=false;
                    break;
                }
            }
            else if(it->x > axis_x){
                double left=axis_x-(it->x - axis_x);
                Point b(left,it->y);
                if(!s.count(b)){
                    good=false;
                    break;
                }
            }
        }
        if(good) puts("YES");
        else puts("NO");
    }
}