UVA - 1595

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

妈蛋这题的测试数据有问题，x 绝对 在【-10000,10000】之外，害得我RE了一天，操

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <vector>
 5 #include <algorithm>
 6 #include <cmath>
 7 using namespace std;
 8 
 9 vector <int> v;
10 vector<vector<int> > Hash(20002);
11 
12 int equal(double x,double y) {
13     return (fabs(x - y) <= 1e-6?1:0);
14 }
15 
16 int main () {
17     int T;
18     // freopen("1.in","r",stdin);
19     cin >> T;
20     while (T--) {
21         int N;
22         cin >> N;
23         for (int i = 0;i < 20002;i++) {
24             Hash[i].clear();
25         }
26         v.clear();
27         int a[20002];
28         memset(a,0,sizeof(a));
29         int x,y;
30         for (int i = 0;i < N;i++) {
31             cin >> x >> y;
32             y += 10000;
33             if (a[y] == 0) {
34                 v.push_back(y);
35                 a[y]++;
36             }
37             Hash[y].push_back(x);
38         }
39         for (int i = 0;i < v.size();i++) {
40             sort(Hash[v[i]].begin(),Hash[v[i]].end());
41         }
42         double mx = ((Hash[v[0]][0] + Hash[v[0]][Hash[v[0]].size() - 1])) * 1.0 / 2;
43         // cout << mx << endl;
44         int ok = 1;
45         for (int i = 0;i < v.size();i++) {
46             if (Hash[v[i]].size() == 1 && !equal(Hash[v[i]][0],mx)) {
47                 ok = 0;break;
48             } 
49             for (int j = 0;j < Hash[v[i]].size() && Hash[v[i]][j] < mx;j++) {
50                 double dis = mx - Hash[v[i]][j];
51                 int xx = Hash[v[i]][j] + int(dis * 2 + 0.5);
52                 if (Hash[v[i]][Hash[v[i]].size() - 1 - j] != xx) {
53                     ok = 0;
54                     break;;
55                 }
56             } 
57             if (ok == 0) break;
58         }
59         cout << (ok ? "YES" : "NO") << endl;
60     }
61 }
View Code

 

posted @ 2014-11-21 15:25  闪光阳  阅读(350)  评论(0)  编辑  收藏  举报