ARC097E. Sorted and Sorted

By solving this problem I better understand the concept of inversion number (aka inversion count). Let $p$ be a permutation of $N$ distinct elements, and $q$ be another permutation of those elements. The inversion number between $p$ and $q$ is defined as the number of ordered pairs of elements $(x, y)$ such that $x$ is to the left of $y$ in permutation $p$, but $x$ is to the right of $y$ in permutation $q$, or equivalently as the number of ordered pairs of elements $(x, y)$ such that $x$ is to the right of $y$ in permutation $p$, but $x$ is to the left of $y$ in permutation $q$.

The inversion number between $p$ and $q$ is the minimum possible number of operations of swapping two adjacent elements to change $p$ into $q$.

Official tutorial:

code

 
int main() {
  int n;
  scan(n);
  vi pos_w(n + 1), pos_b(n + 1);
  rng (i, 0, 2 * n) {
    char color;
    int number;
    scan(color, number);
    if (color == 'B') {
      pos_b[number] = i;
    } else {
      pos_w[number] = i;
    }
  }
  // nb_after[i][j] : number of black balls placed after position i numbered between 1 and j.
  // nw_after[i][j] : number of white balls placed after position i numbered between 1 and j.
  vv nb_after(2 * n, vi(n + 1)), nw_after(2 * n, vi(n + 1));
  rng (i, 0, 2 * n) {
    up (j, 1, n) {
      nb_after[i][j] = nb_after[i][j - 1] + (pos_b[j] > i);
      nw_after[i][j] = nw_after[i][j - 1] + (pos_w[j] > i);
    }
  }
  // dp[i][j] : minimum possible inversion number among first i black balls and first j white balls
  vv dp(n + 1, vi(n + 1));
  up (i, 1, n) {
    dp[i][0] = dp[i - 1][0] + nb_after[pos_b[i]][i - 1];
  }
  up (i, 1, n) {
    dp[0][i] = dp[0][i - 1] + nw_after[pos_w[i]][i - 1];
  }
  up (i, 1, n) {
    up (j, 1, n) {
      dp[i][j] = min(dp[i][j - 1] + nb_after[pos_w[j]][i] + nw_after[pos_w[j]][j - 1],
                     dp[i - 1][j] + nb_after[pos_b[i]][i - 1] + nw_after[pos_b[i]][j]);
    }
  }
  println(dp[n][n]);
  return 0;
}