Problem:
Score : 600 points

Problem Statement

There are 2N balls, N white and N black, arranged in a row. The integers from 1 through N are written on the white balls, one on each ball, and they are also written on the black balls, one on each ball.
The integer written on the i-th ball from the left (1 ≤ i ≤ 2N) is a_i, and the color of this ball is represented by a letter c_i.
c_i = W represents the ball is white; c_i = B represents the ball is black.

Takahashi the human wants to achieve the following objective:

For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the white ball with i written on it is to the left of the white ball with j written on it.

For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the black ball with i written on it is to the left of the black ball with j written on it.

In order to achieve this, he can perform the following operation:

Swap two adjacent balls.

Find the minimum number of operations required to achieve the objective.

Constraints

1 ≤ N ≤ 2000

1 ≤ a_i ≤ N

c_i = W or c_i = B.

If i ≠ j, (a_i,c_i) ≠ (a_j,c_j).

Input

Input is given from Standard Input in the following format:

N
c_1 a_1
c_2 a_2
:
c_{2N} a_{2N}

Output

Print the minimum number of operations required to achieve the objective.

Sample Input 1

3
B 1
W 2
B 3
W 1
W 3
B 2

Sample Output 1

4

The objective can be achieved in four operations, for example, as follows:

Swap the black 3 and white 1.

Swap the white 1 and white 2.

Swap the black 3 and white 3.

Swap the black 3 and black 2.

Sample Input 2

4
B 4
W 4
B 3
W 3
B 2
W 2
B 1
W 1

Sample Output 2

18

Sample Input 3

9
W 3
B 1
B 4
W 1
B 5
W 9
W 2
B 6
W 5
B 3
W 8
B 9
W 7
B 2
B 8
W 4
W 6
B 7

Sample Output 3

41

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)


#include<iostream>
#include<cstdio>
#include<cstring>
#include <cstdlib>  
#include <math.h>
#include <cmath>
#include<cctype>
#include<string>
#include<set>
#include<iomanip>
#include <map>
#include<algorithm>
#include <functional>
#include<vector>
#include<climits>
#include<stack>
#include<queue>
#include<bitset>
#include <deque>
#include <climits>
#include <typeinfo>
#include <utility> 
using namespace std;
using ll = long long;
template<typename T>using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const ll inf = 1145141919810364364LL;
#define all(x) (x).begin(),(x).end()
#define puts(x) cout << x << endl
#define rep(i,m,n) for(ll i = m;i < n;++i)
#define pb push_back
#define fore(i,a) for(auto &i:a)
#define rrep(i,m,n) for(ll i = m;i >= n;--i)

template<typename T>
struct BIT {
	int n;
	vector<T>dat;
	BIT(int n = 0) {
		initialize(n);
	}
	void initialize(int temp) {
		n = temp;
		dat.resize(n);
		rep(i, 0, n)dat[i] = 0;
	}
	T sum(int i) {
		T s = 0;
		while (i >= 0) {
			s += dat[i];
			i = (i & (i + 1)) - 1;
		}
		return s;
	}
	T sum_between(int i, int j) {
		if (i > j)return 0;
		return sum(j) - sum(i - 1);
	}
	void plus(int i, T x) {
		while (i < n) {
			dat[i] += x;
			i |= i + 1;
		}
	}
	int lower_bound(T x) {
		if (sum(n - 1) < x)return -1;
		else if (sum(0) >= x)return 0;
		int ng = 0, ok = n - 1;
		while (ok - ng > 1) {
			int mid = (ok + ng) / 2;
			if (sum(mid) >= x)ok = mid;
			else ng = mid;
		}
		return ok;
	}
};
ll dp[2020][2020];
vector<pair<char,int>>v(4040);
map<pair<char, int>, int>mp;
int cntw[2020][2020],cntb[2020][2020];


int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int N;
	cin >> N;
	for (int i = 0; i < 2 * N; i++) {
		cin >> v[i].first >> v[i].second;
		mp[{v[i].first,v[i].second}] = i;
	}
	for (int i = 0; i < 2020; i++) {
		for (int j = 0; j < 2020; j++) {
			dp[i][j] = inf;
		}
	}

	BIT<ll>bitw(2020), bitb(2020);
	for (int i = 2 * N - 1; i >= 0; i--) {
		for (int j = 1; j <= N; j++) {
			cntw[i][j] = bitw.sum_between(0,j);
			cntb[i][j] = bitb.sum_between(0,j);
		}
		if (v[i].first == 'W')bitw.plus(v[i].second,1);
		else bitb.plus(v[i].second,1);
	}
	
	dp[0][0] = 0;
	for (int i = 0; i <= N; i++) {
		for (int j = 0; j <= N; j++){
			ll cost1 = cntw[mp[{'W',i+1}]][i] + cntb[mp[{'W',i+1}]][j];
			ll cost2 = cntw[mp[{'B',j+1}]][i] + cntb[mp[{'B',j+1}]][j];
			dp[i+1][j] = min(dp[i+1][j],dp[i][j]+cost1);
			dp[i][j+1] = min(dp[i][j+1],dp[i][j]+cost2);
		}
	}
	puts(dp[N][N]);

	return 0;
}