題目

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.

Paste_Image.png

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

分析

給你一堆點(diǎn)，讓你看能不能找到一條豎線使所有點(diǎn)左右對(duì)稱(chēng)起趾。思路很簡(jiǎn)單瓣颅，開(kāi)一個(gè)20000的vector存點(diǎn)然后對(duì)每一行找對(duì)稱(chēng)點(diǎn)葬项，看看是不是在一條豎線上，額我寫(xiě)的有點(diǎn)取巧簿寂，其實(shí)是有樣例可以卡掉我的代碼，但是竟然過(guò)了潮饱，蝉衣，括尸，
所以我又寫(xiě)了個(gè)正確的代碼，病毡，

ac代碼

//ac代碼1濒翻，錯(cuò)誤代碼
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
const int maxn = 2e4 + 10;
const int mid = 1e4 + 5;
vector <int> dot[maxn];
int n;
void init(){
    for(int i = 0; i < maxn; ++i){
        dot[i].clear();
    }
    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
        int t1, t2;
        scanf("%d%d", &t1, &t2);
        dot[t2 + mid].push_back(t1);
    }
}
void solve(){
    bool text = 1;
    double t;
    for(int i = 0; i < maxn; ++i){
        if(dot[i].size() == 0) continue;
        double sum = 0;
        for(int j = 0; j < dot[i].size(); ++j){
            sum += dot[i][j];
        }
        sum /= dot[i].size();
        if(text){
            t = sum;
            text = 0;
        }
        else {
            if(t != sum){
                printf("NO\n");
                return;
            }
        }
    }
    printf("YES\n");
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        init();
        solve();
    }
    return 0;
}

//ac代碼2，正確（暫時(shí)沒(méi)發(fā)現(xiàn)什么錯(cuò)誤
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
const int maxn = 2e4 + 10;
const int mid = 1e4 + 5;
vector <int> dot[maxn];
int n;
void init(){
    for(int i = 0; i < maxn; ++i){
        dot[i].clear();
    }
    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
        int t1, t2;
        scanf("%d%d", &t1, &t2);
        dot[t2 + mid].push_back(t1);
    }
}
void solve(){
    bool text = 1;
    double t;
    for(int i = 0; i < maxn; ++i){
        if(dot[i].size() == 0) continue;
        double sum = 0;
        for(int j = 0; j < dot[i].size(); ++j){
            sum += dot[i][j];
        }
        sum /= dot[i].size();
        if(text){
            t = sum;
            text = 0;
        }
        else {
            if(t != sum){
                printf("NO\n");
                return;
            }
        }
    }
    for(int i = 0; i < maxn; ++i){
        for(int j = 0; j < dot[i].size(); ++j){
            if(dot[i][j] == t){
                continue;
            } 
            else{
                int temp = t * 2 - dot[i][j];
                for(int k = 0; ; ++k){
                    if(k == dot[i].size()){
                        printf("NO\n");
                        return;
                    }
                    else if(dot[i][k] == temp) break;
                }
            }
        }
    }
    printf("YES\n");
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        init();
        solve();
    }
    return 0;
}