【HNOI2016】网络

题面

题解

考虑整体二分。

定义整体二分函数solve(l, r, ql, qr)表示操作权值在\([l, r]\)中，对\([ql, qr]\)的询问进行二分。

这样的话check就会很简单，先按照时间将所有\(\geq mid\)的边加进去，对于每个点判断是不是所有路径都经过了这个点就可以判断这个点的答案是不是\(\geq mid\)

具体如何判断的话可以用树上差分。

代码

#include<cstdio>

#include<cstring>

#include<cctype>

#include<algorithm>

#define RG register

#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)

#define clear(x, y) memset(x, y, sizeof(x))

inline int read()

{

	int data = 0, w = 1; char ch = getchar();

	while(ch != '-' && (!isdigit(ch))) ch = getchar();

	if(ch == '-') w = -1, ch = getchar();

	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();

	return data * w;

}

const int maxn(100010), maxm(200010);

struct edge { int next, to; } e[maxn << 1];

int head[maxn], e_num, n, m;

inline void add_edge(int from, int to)

{

	e[++e_num] = (edge) {head[from], to};

	head[from] = e_num;

}

int size[maxn], heavy[maxn], pos[maxn], fa[maxn], belong[maxn], cnt;

void dfs(int x)

{

	size[x] = 1;

	for(RG int i = head[x]; i; i = e[i].next)

	{

		int to = e[i].to; if(to == fa[x]) continue;

		fa[to] = x; dfs(to); size[x] += size[to];

		if(size[heavy[x]] < size[to]) heavy[x] = to;

	}

}

void dfs(int x, int chain)

{

	pos[x] = ++cnt; belong[x] = chain;

	if(!heavy[x]) return;

	dfs(heavy[x], chain);

	for(RG int i = head[x]; i; i = e[i].next)

	{

		int to = e[i].to;

		if(to == fa[x] || to == heavy[x]) continue;

		dfs(to, to);

	}

}

struct qry { int opt, from, to, dis, id, lca, y; } s[maxm], pl[maxm], pr[maxm];

int U[maxm], ucnt, ans[maxm], c[maxm];

void update(int x, int v) { while(x <= n) c[x] += v, x += x & -x; }

int query(int x) { int a = 0; while(x) a += c[x], x -= x & -x; return a; }

void Div(int l, int r, int ql, int qr)

{

	if(ql > qr) return;

	bool flag = 1; int cntl = 0, cntr = 0;

	for(RG int i = ql; i <= qr; i++)

		if(s[i].opt == 2) { flag = 0; break; }

	if(flag) return;

	if(l == r)

	{

		for(RG int i = ql; i <= qr; i++)

			if(s[i].id) ans[s[i].id] = l;

		return;

	}

	int mid = (l + r) >> 1, sum = 0;

	for(RG int i = ql; i <= qr; i++)

		if(s[i].opt == 2)

		{

			if(query(pos[s[i].from] + size[s[i].from] - 1) -

				query(pos[s[i].from] - 1) == sum) pr[++cntr] = s[i];

			else pl[++cntl] = s[i];

		}

		else if(s[i].dis <= mid)

		{

			int d = s[i].opt ? -1 : 1; sum += d;

			update(pos[s[i].from], d); update(pos[s[i].to], d);

			update(pos[s[i].lca], -d);

			if(s[i].lca != 1) update(pos[fa[s[i].lca]], -d);

			pl[++cntl] = s[i];

		}

		else pr[++cntr] = s[i];

	memcpy(s + ql, pl + 1, sizeof(qry) * cntl);

	memcpy(s + ql + cntl, pr + 1, sizeof(qry) * cntr);

	for(RG int i = ql; i <= qr; i++)

		if(s[i].opt != 2 && s[i].dis <= mid && s[i].y)

		{

			int d = s[i].opt ? 1 : -1;

			update(pos[s[i].from], d); update(pos[s[i].to], d);

			update(pos[s[i].lca], -d);

			if(s[i].lca != 1) update(pos[fa[s[i].lca]], -d);

		}

	Div(l, mid, ql, ql + cntl - 1), Div(mid + 1, r, ql + cntl, qr);

}

int main()

{

	n = read(), m = read();

	for(RG int i = 1, a, b; i < n; i++)

		a = read(), b = read(),

		add_edge(a, b), add_edge(b, a);

	dfs(1); dfs(1, 1);

	for(RG int i = 1; i <= m; i++)

	{

		s[i].opt = read();

		if(s[i].opt == 0)

		{

			int x = s[i].from = read(), y = s[i].to = read();

			U[++ucnt] = -(s[i].dis = read()), s[i].y = 1;

			while(belong[x] != belong[y])

			{

				if(pos[belong[x]] < pos[belong[y]]) std::swap(x, y);

				x = fa[belong[x]];

			}

			s[i].lca = (pos[x] < pos[y] ? x : y);

		}

		else if(s[i].opt == 1)

		{

			int x = read(); s[i] = s[x]; s[i].opt = 1;

			s[i].y = s[x].y = 0;

		}

		else s[i].from = read(), s[i].id = ++ans[0];

	}

	U[++ucnt] = 1; std::sort(U + 1, U + ucnt + 1);

	ucnt = std::unique(U + 1, U + ucnt + 1) - U - 1;

	for(RG int i = 1; i <= m; i++) if(s[i].opt != 2)

		s[i].dis = std::lower_bound(U + 1, U + ucnt + 1, -s[i].dis) - U;

	for(RG int i = 1; i <= ucnt; i++) U[i] = -U[i];

	Div(1, ucnt, 1, m);

	for(RG int i = 1; i <= ans[0]; i++) printf("%d\n", U[ans[i]]);

	return 0;

}

巴特西

【HNOI2016】网络

题面

题解

代码

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