Ball Coloring

There are N bags, each containing two white balls. The i-th box contains two balls with integers xi and yi written on them, respectively.
For each of these bags, you will paint one of the balls red, and paint the other blue.
Afterwards, the 2N balls will be classified according to color.
Then, we will define the following:
Rmax: the maximum integer written on a ball painted in red
Rmin: the minimum integer written on a ball painted in red
Bmax: the maximum integer written on a ball painted in blue
Bmin: the minimum integer written on a ball painted in blue
Find the minimum possible value of (Rmax?Rmin)×(Bmax?Bmin).

Constraints
1≤N≤200,000
1≤xi,yi≤109

Input is given from Standard Input in the following format:
N
x1 y1
x2 y2
:
xN yN

Print the minimum possible value.

3
1 2
3 4
5 6

15

The optimal solution is to paint the balls with x1, x2, y3 red, and paint the balls with y1, y2, x3 blue.

思维题，解法：

考虑所有数中最大的和最小的：MIN，MAXMIN，MAX，它们要么在两个集合中，要么只在一个集合中。不失一般性，我们考虑以下两种情况： 
　　 
　　1．Rmin=MIN，Bmax=MAXRmin=MIN，Bmax=MAX 
　　这时，我们要最小化RmaxRmax，最大化BminBmin，那么就只要在分组时把小的分给RR，大的分给BB即可。 
　　 
　　2．Rmin=MIN，Rmax=MAXRmin=MIN，Rmax=MAX 
　　这时，只需最小化Bmax?BminBmax?Bmin。这样的话我们知道肯定是要取，比如说2n2n个数一起排完序后，中间连续互斥的nn个。这样的话，我们先让每组中的xi<yixi<yi，然后按照xx数组排序，先让BB数组取x1～xnx1～xn，然后再逐个把xixi调成yiyi计算答案即可。

AC代码：

#include <bits/stdc++.h>
using namespace std;
const long long INF=1e18+10;
typedef pair<long long,long long>PA;
vector<PA> V;
multiset<long long>R,B;
int cmp(PA a,PA b)
{
return a.first<b.first;
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int n;
long long x,y,ans;
cin>>n;
for(int i=1; i<=n; i++)
{
cin>>x>>y;
V.push_back(make_pair(min(x,y),max(x,y)));
R.insert(min(x,y));
B.insert(max(x,y));
}
long long maR=-INF,maB=-INF;
long long miR=INF,miB=INF;
sort(V.begin(),V.end(),cmp);
for(int i=0; i<V.size(); i++)
{
maR=max(V[i].first,maR);
maB=max(V[i].second,maB);
miR=min(V[i].first,miR);
miB=min(V[i].second,miB);
}
ans=(maR-miR)*(maB-miB);
for(int i=0; i<V.size(); i++)
{
R.erase(R.find(V[i].first));
B.insert(V[i].first);
B.erase(B.find(V[i].second));
R.insert(V[i].second);
ans=min(ans,(*R.rbegin()-*R.begin())*(*B.rbegin()-*B.begin()));
}
cout<<ans<<endl;
return 0;
}