bzoj 4849 Mole Tunnels

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模拟费用流.

模拟费用流是怎么回事呢?模拟相信大家都很熟悉,但是模拟费用流是怎么回事呢,下面就让小编带大家一起了解吧.

模拟费用流,其实就是模拟费用流的过程,大家可能会很惊讶,模拟怎么会费用流呢?

但事实就是这样,小编也感到非常惊讶.

这就是关于模拟费用流的事情了,大家有什么想法呢,欢迎在评论区告诉小编一起讨论哦!

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//%std
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
	int out = 0, fh = 1;
	char jp = getchar();
	while ((jp > '9' || jp < '0') && jp != '-')
		jp = getchar();
	if (jp == '-')
		fh = -1, jp = getchar();
	while (jp >= '0' && jp <= '9')
		out = out * 10 + jp - '0', jp = getchar();
	return out * fh;
}
const int N = 1e5 + 10, inf = 1e9;
int n, m, c[N], ans = 0, idx = 0, st[N], ed[N], rnk[N], dep[N];
void dfs(int x)
{
	st[x] = ++idx, rnk[idx] = x;
	if (2 * x <= n)
	{
		dep[2 * x] = dep[x] + 1;
		dfs(2 * x);
	}
	if (2 * x + 1 <= n)
	{
		dep[2 * x + 1] = dep[x] + 1;
		dfs(2 * x + 1);
	}	
	ed[x] = idx;
}
struct node
{
	int mi, mi_real, tag, pos;
} Tree[N << 2];
#define root Tree[o]
#define lson Tree[o << 1]
#define rson Tree[o << 1 | 1]
void pushup(int o)
{
	root.mi = min(lson.mi, rson.mi);
	root.mi_real = min(lson.mi_real, rson.mi_real);
	if (lson.mi < rson.mi)
		root.pos = lson.pos;
	else
		root.pos = rson.pos;
}
void BuildTree(int o, int l, int r)
{
	if (l == r)
	{
		root.mi = c[rnk[l]] ? dep[rnk[l]] : inf;
		root.mi_real = dep[rnk[l]];
		root.pos = rnk[l];
		return;
	}
	int mid = (l + r) >> 1;
	BuildTree(o << 1, l, mid);
	BuildTree(o << 1 | 1, mid + 1, r);
	pushup(o);
}
void modify(int o, int c)
{
	root.mi += c;
	if (c < inf)
		root.mi_real += c;
	root.tag += c;
}
void pushdown(int o)
{
	if (root.tag)
	{
		modify(o << 1, root.tag);
		modify(o << 1 | 1, root.tag);
		root.tag = 0;
	}
}
void upd(int o, int l, int r, int L, int R, int c)
{
	if (L <= l && r <= R)
		return modify(o, c);
	int mid = (l + r) >> 1;
	pushdown(o);
	if (L <= mid) 
		upd(o << 1, l, mid, L, R, c);
	if (R > mid)
		upd(o << 1 | 1, mid + 1, r, L, R, c);
	pushup(o);
}
void query(int &pos, int &mi, int o, int l, int r, int L, int R)
{
	if (L <= l && r <= R)
	{
		if (root.mi < mi)
			mi = root.mi, pos = root.pos;
		return;
	}
	pushdown(o);
	int mid = (l + r) >> 1;
	if (L <= mid)
		query(pos, mi, o << 1, l, mid, L, R);
	if (R > mid)
		query(pos, mi, o << 1 | 1, mid + 1, r, L, R);
}
int query_real(int o, int l, int r, int pos)
{
	if (l == r)
		return root.mi_real;
	pushdown(o);
	int mid = (l + r) >> 1;
	if (pos <= mid)
		return query_real(o << 1, l, mid, pos);
	else
		return query_real(o << 1 | 1, mid + 1, r, pos);
}
int up[N], down[N]; // available flow of inverse edge
int main()
{
	n = read(), m = read();
	dfs(1);
	for (int i = 1; i <= n; ++i)
		c[i] = read();
	BuildTree(1, 1, n);
	while (m--)
	{
		int x = read(), lca = x, mi, tmp, d = 0, dis = inf, y, p;
		while (lca)
		{
			mi = inf;
			query(tmp, mi, 1, 1, n, st[lca], ed[lca]);
			if (mi < inf)
			{
				int k = d + mi - query_real(1, 1, n, st[lca]);
				if (k < dis)
					dis = k, y = tmp, p = lca;
			}
			d += up[lca] ? -1 : 1;
			lca >>= 1;
		}
		ans += dis;
		printf("%d ", ans);

		--c[y];
		if (!c[y])
			upd(1, 1, n, st[y], st[y], inf);
		lca = p;
		
		// x -> lca
		p = x;
		while (p != lca)
		{
			if (up[p]) // used inverse edge
				--up[p];
			else // not
			{
				++down[p];
				if (down[p] == 1) // cost : 1 -> -1
					upd(1, 1, n, st[p], ed[p], -2);
			}
			p >>= 1;
		}
		
		// lca -> y
		p = y;
		while (p != lca)
		{
			if (down[p]) // used inverse edge
			{
				--down[p];
				if (down[p] == 0) // cost : -1 -> 1
					upd(1, 1, n, st[p], ed[p], 2);
			}	
			else // not
				++up[p];
			p >>= 1;
		}
	}
	return 0;
}