【luogu P3128 [USACO15DEC]最大流Max Flow】 题解

题目链接：https://www.luogu.org/problemnew/show/P3128
菜

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define lson left, mid, rt<<1
#define rson mid+1, right, rt<<1|1
using namespace std;
const ll maxn = 100000 + 10;
ll n, m, root;
ll fa[maxn], top[maxn], son[maxn], size[maxn], rev[maxn], seg[maxn], deep[maxn];
ll lazy[maxn<<2], tree[maxn<<2], res, num;
struct edge{
	ll from, to, next;
}e[maxn<<2];
ll head[maxn], cnt;
void add(ll u, ll v)
{
	e[++cnt].from = u;
	e[cnt].to = v;
	e[cnt].next = head[u];
	head[u] = cnt;
}
//-----segment_tree-----
void PushUP(ll rt)
{
	tree[rt] = max(tree[rt<<1], tree[rt<<1|1]);
}
void build(ll left, ll right, ll rt)
{
	if(left == right)
	{
		tree[rt] = rev[left];
		return;
	}
	ll mid = (left + right)>>1;
	build(lson);
	build(rson);
	PushUP(rt);
}
void PushDOWN(ll left, ll right, ll rt, ll mid)
{
	lazy[rt<<1] += lazy[rt];
	lazy[rt<<1|1] += lazy[rt];
	tree[rt<<1] += lazy[rt];//区间求和才需要加(mid - left + 1)*lazy[rt]
	tree[rt<<1|1] += lazy[rt];//区间求最值只需要求一个值只需要加一次lazy[rt] 
	lazy[rt] = 0; 
}
void update(ll l, ll r, ll add, ll left, ll right, ll rt)
{
	if(l <= left && r >= right)
	{
		tree[rt] += add;
		lazy[rt] += add;
		return;
	}
	ll mid = (left + right)>>1;
	if(lazy[rt]) PushDOWN(left, right, rt, mid);
	if(l <= mid) update(l, r, add, lson);
	if(r > mid) update(l, r, add, rson);
	PushUP(rt);
}
ll query(ll l, ll r, ll left, ll right, ll rt)
{
	ll res = -0x7fffffff;
	if(l <= left && r >= right)
	{
		return tree[rt];
	}
	ll mid = (left + right)>>1;
	if(lazy[rt]) PushDOWN(left, right, rt, mid);
	if(l <= mid) res = max(query(l, r, lson), res);
	if(r > mid) res = max(query(l, r, rson), res);
	return res;
}
//----------------------
void dfs1(ll u, ll f, ll d)
{
	ll maxson = -1;
	size[u] = 1;
	deep[u] = d;
	fa[u] = f;
	for(ll i = head[u]; i != -1; i = e[i].next)
	{
		ll v = e[i].to;
		if(f != v)
		{
			dfs1(v, u, d+1);
			size[u] += size[v];
			if(maxson < size[v]) son[u] = v, maxson = size[v];
		}
	}
}
void dfs2(ll u, ll t)
{
	seg[u] = ++num;
	rev[num] = 0;
	top[u] = t;
	if(!son[u]) return;
	dfs2(son[u], t);
	for(ll i = head[u]; i != -1; i = e[i].next)
	{
		ll v = e[i].to;
		if(v == fa[u] || v == son[u]) continue;
		dfs2(v,v);
	}
}
void updRange(ll x, ll y)
{
    while(top[x] != top[y])
    {
        if(deep[top[x]] < deep[top[y]]) swap(x, y);
        update(seg[top[x]], seg[x], 1, 1, n, 1);
        x = fa[top[x]];
    }
    if(deep[x] > deep[y]) swap(x, y);
    update(seg[x], seg[y], 1, 1, n, 1);
}
int main()
{
	memset(head, -1, sizeof(head));
	scanf("%lld%lld",&n,&m); root = 1;
	for(ll i = 1; i < n; i++) 
	{
		ll u, v;
		scanf("%lld%lld",&u,&v);
		add(u, v), add(v, u);
	}
	dfs1(root, 0, 1);
	dfs2(root, root);
	build(1, n, 1);
	for(ll i = 1; i <= m; i++)
	{
		ll u, v;
		scanf("%lld%lld",&u,&v);
		updRange(u, v);
	}
	printf("%lld",tree[1]);
}