CQOI 2011 动态逆序对 【CDQ分治】

CDQ分治与普通分治不同的点在于CDQ只统计当前区间的左右连个子区间之间的贡献，而不是各个区间内部的贡献

这题先把逆序对求出来，然后计算出每次删除的数会减少多少逆序对即可，把被删除的数字的时间也当成一位偏序，离线做，这样使得只有在一个点之后被删除的点才会对这个点造成贡献

注意每次删除给的不是点的下标，而是值

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define gi get_int()
#define int long long
const int MAXN = 2e6;
int get_int()
{
  int x = 0, y = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-')
    ch = getchar();
  if (ch == '-')
    y = -1, ch = getchar();
  while (isdigit(ch))
    x = x * 10 + ch - '0', ch = getchar();
  return x * y;
}

int tree[MAXN], count[2][MAXN], quest[MAXN], map[MAXN];

int lowbit(int u)
{
  return u & (-u);
}

void insert(int index, int value)
{
  for (; index < MAXN; index += lowbit(index))
    tree[index] += value;
}

int query(int index)
{
  int ans = 0;
  for (; index > 0; index -= lowbit(index))
    ans += tree[index];
  return ans;
}

class Node
{
  public:
    int index, v, time;
} num[MAXN];
int cmp1(const Node& a, const Node& b)
{
  if (a.v == b.v)
    return a.index < b.index;
  return a.v > b.v;
}
int cmp2(const Node& a, const Node& b)
{
  if (a.v == b.v)
    return a.index < b.index;
  return a.v < b.v;
}
int cmp3(const Node& a, const Node& b)
{
  return a.index > b.index;
}

void CDQ(int l, int r, int mode)
{
  if (l == r - 1)
    return ;
  int mid = (l + r) / 2;
  CDQ(l, mid, mode);
  CDQ(mid, r, mode);
  if (mode == 0) {
    std::sort(num + l, num + mid, cmp1);
    std::sort(num + mid, num + r, cmp1);
  } else {
    std::sort(num + l, num + mid, cmp2);
    std::sort(num + mid, num + r, cmp2);
  }
  int p1 = l, p2 = mid;
  while (p1 < mid && p2 < r) {
    if ((mode == 0 && num[p1].v >= num[p2].v) || (mode == 1 && num[p1].v <= num[p2].v)) {
      insert(num[p1].time, 1);
      p1++;
    } else {
      count[mode][num[p2].index] += query(num[p2].time);
      p2++;
    }
  }
  while (p2 < r) {
    count[mode][num[p2].index] += query(num[p2].time);
    p2++;
  }
  while (p1 > l) {
    p1--;
    insert(num[p1].time, -1);
  }
}

signed main()
{
  freopen("code.in", "r", stdin);
  freopen("code.out", "w", stdout);
  int n = gi, m = gi;
  for (int i = 0; i < n; i++) {
    num[i].index = i;
    num[i].v = gi;
    map[num[i].v] = i;
    num[i].time = 1;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    ans += query(n + 1 - num[i].v);
    insert(n + 1 - num[i].v, 1);
  }
  memset(tree, 0, sizeof(tree));
  int tmp = n + 1;
  for (int i = 0; i < m; i++) {
    int index = gi;
    quest[i] = index;
    num[map[index]].time = tmp--;
  }
  CDQ(0, n, 0);
  std::sort(num, num + n, cmp3);
  CDQ(0, n, 1);
  std::cout << ans << std::endl;
  for (int i = 0; i < m - 1; i++) {
    int now = map[quest[i]];
    ans -= count[0][now] + count[1][now];
    std::cout << ans << std::endl;
  }

  return 0;
}