UVA_1595: Symmetry

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Description

The figure shown on the left is left-right symmetric as it is possible to fold the sheet of paper along avertical 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 casesT is given in the first line of the input file. The first line of each test case contains an integerN , 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 andy-coordinate of a dot. Both x-coordinatesand 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 6 105 10 6 14

Sample Output

YES NO YES

分析：分别对每行，即y相同的点求对称轴，即x的值，然后比较每行的轴是否一致即可；注意每行的轴即每两个点的轴，中间循环判断点之间的轴是否相等，每行处理前对x排序

#include <iostream>#include <vector>#include <algorithm>#include <map>#include <cstdio>#include <cmath>using namespace std;const int maxn = 1000 + 10;typedef vector<int> pointx;vector<pointx> dot;map<int,int> IDcache;void print(){for(int i=0; i<dot.size(); i++){for(int j=0; j<dot[i].size(); j++){printf("%4d",dot[i][j]);}putchar('\n');}}int main(){int T,n;cin >> T;while(T--){        cin >> n;int x,y;dot.clear(); IDcache.clear();        for(int i=0; i<n; i++)        {cin >> x >> y;if(IDcache.count(y)){                int index = IDcache[y];                dot[index].push_back(x);}else{pointx dotx;dotx.push_back(x);dot.push_back(dotx);IDcache[y]=dot.size()-1;}        }int yes = 1;int ave;int avebefore;        for(int i=0; i<dot.size(); i++)        {sort(dot[i].begin(),dot[i].end());int dotisize = dot[i].size();if(dotisize%2)ave=dot[i][dotisize/2]*2;else {ave=dot[i][0]+dot[i][dotisize-1];}if(i==0)avebefore=ave;else if(avebefore!=ave){yes=0;break;}int j=0;for(j=0; j<dotisize/2; j++){int avej=dot[i][j]+dot[i][dotisize-j-1];if(avej!=ave){yes=0;break;}}        }printf("%s\n",yes?"YES":"NO");}    return 0;}

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