【bzoj3295】[Cqoi2011]动态逆序对

题目描述：

对于序列A，它的逆序对数定义为满足i<j，且Ai>Aj的数对(i,j)的个数。给1到n的一个排列，按照某种顺序依次删除m个元素，你的任务是在每次删除一个元素之前统计整个序列的逆序对数。

输入：

输入第一行包含两个整数n和m，即初始元素的个数和删除的元素个数。以下n行每行包含一个1到n之间的正整数，即初始排列。以下m行每行一个正整数，依次为每次删除的元素。

输出：
输出包含m行，依次为删除每个元素之前，逆序对的个数。

样例输入：
5 4
1
5
3
4
2
5
1
4
2

样例输出：
5
2
2
1

题解：
动态逆序对，如果把逆序对当做二维偏序，动态的相当于多一维时间。于是就变成了三维偏序问题，然后就可以用cdq来做了。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

#ifdef WIN32
	#define LL "%I64d"
#else
	#define LL "%lld"
#endif

#ifdef CT
	#define debug(...) printf(__VA_ARGS__)
	#define setfile()
#else
	#define debug(...)
	#define filename ""
	#define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);
#endif

#define R register
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))
#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))
#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)
#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)
char B[1 << 15], *S = B, *T = B;
inline int FastIn()
{
	R char ch; R int cnt = 0; R bool minus = 0;
	while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;
	ch == '-' ? minus = 1 : cnt = ch - '0';
	while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';
	return minus ? -cnt : cnt;
}
#define maxn 100010
#define maxm 50010
int pos[maxn], bit[maxn], last[maxn], now, n, m;
struct Event
{
	int pos, t, val;
	inline bool operator < (const Event &that) const
	{
		return t < that.t || (t == that.t && (pos < that.pos || (pos == that.pos && val < that.val)));
	}
}p[maxn], t[maxn];
int ans[maxn];
#define lowbit(_x) ((_x) & -(_x))
inline void add(R int x, R int val)
{
	for (; x <= n; x += lowbit(x))
	{
		if (last[x] != now)
			bit[x] = 0;
		last[x] = now;
		bit[x] += val;
	}
}
inline int query(R int x)
{
	R int ret = 0;
	for (; x ; x -= lowbit(x))
		if (last[x] == now)
			ret += bit[x];
	return ret;
}
void cdq(R int left, R int right)
{
	if (left == right) return ;
	R int mid = left + right >> 1;
	R int i, j, k;
	for (i = k = left, j = mid + 1; k <= right; ++k)
		t[p[k].t <= mid ? i++ : j++] = p[k];
	for (R int i = left; i <= right; ++i)
		p[i] = t[i];
	++now;
	for (R int i = left, j = mid + 1; j <= right; ++j)
	{
		for (; i <= mid && p[i].pos <= p[j].pos; ++i)
			add(p[i].val, 1);
		ans[p[j].t] += i - left - query(p[j].val);
	}
	++now;
	for (R int i = mid, j = right; j > mid; --j)
	{
		for (; i >= left && p[i].pos >= p[j].pos; --i)
			add(p[i].val, 1);
		ans[p[j].t] += query(p[j].val);
	}
	cdq(left, mid); cdq(mid + 1, right);
}
long long sum[maxn];
int main()
{
//	setfile();
	n = FastIn(); m = FastIn();
	for (R int i = 1; i <= n; ++i)
	{
		R int val = FastIn();
		p[i] = (Event) {i, 0, val}, pos[val] = i;
	}
	for (R int i = 1; i <= m; ++i)
		p[pos[FastIn()]].t = n - i + 1;
	R int mcnt = 0;
	for (R int i = 1; i <= n; ++i)
		if (!p[i].t) p[i].t = ++mcnt;
	cdq(1, n);
	for (R int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + ans[p[i].t];
	for (R int i = n; i > n - m; --i) printf("%lld\n",sum[i] );
	return 0;
}
/*
5 4
1 5 3 4 2
5
1
4
2
*/