joj 2197 perfect set 序偶对称

joj 2197 perfect set 序偶对称

A perfect point set is a set of points, for every point (x,y) in this set, point (y,x) is also in this set.

Input

There are mutical cases for this problem. For every input case, there is a set. we give you the number of points in this set in the first line, followed by n lines,each give two integers x y.

Output

If this set is a perfect set, you should print "Yes", else you print "No".

Sample Input

4
1 4
4 1
2 5
5 2

Sample Output

Yes

1，这种类似稀疏矩阵的数组对一般不用数组存储，用结构体比较好

2，排序后比较非常的巧妙。

#include<iostream>
#include<cstdlib>
using namespace std;
struct mm{
double a;
double b;}M[100000];
bool  operator<(mm m1,mm m2)
{
if(m1.a<m2.a)
return true;
else
{
if(m1.a==m2.a&&m1.b<m2.b)
return true;
}
return false;
}
int main()
{
// freopen("s.txt","r",stdin);
// freopen("key.txt","w",stdout);
int num;
double a,b;
int temp=0;
while(cin>>num)
{
memset(M,0,sizeof(M));
temp=0;
for(int k=0;k<num;k++)
{
cin>>a>>b;
if(a<b)
{
M[temp].a=a;
M[temp].b=b;
temp++;
}
else if(a>b)//把a和b相等的都去掉了
{
M[temp].a=b;
M[temp].b=a;
temp++;
}
}
if(temp%2!=0)cout<<"No"<<endl;
else
{
sort(M,M+temp);
int flag=0;
for(int p=0;p<temp;p+=2)
{
if(M[p].b!=M[p+1].b||M[p].a!=M[p+1].a)
{
flag=1;break;
}
}
if(flag==0)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
}
//system("PAUSE");
return   0;
}