【luogu P4114 Qtree1】 题解

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题目链接：https://www.luogu.org/problemnew/show/P4114
1.把边权转化到点权：选取连接这条边的两个点中较深的一个。
2.查询点到点之间的边权时，要从seg[x]+1 到 seg[y]，因为seg[x]其实连接的是上面一条边的边权。

#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 = 300000 + 10;
char opt[9];
ll n, m, root, res;
ll top[maxn], rev[maxn], seg[maxn], son[maxn];
ll fa[maxn], size[maxn], deep[maxn];
ll u[maxn], v[maxn], w[maxn];
ll tree[maxn<<2], lazy[maxn<<2], num, node[maxn];
struct edge{
    ll next, from, to, len;
}e[maxn<<2];
ll head[maxn], cnt;
void add(ll u, ll v, ll w)
{
    e[++cnt].from = u;
    e[cnt].len = w;
    e[cnt].next = head[u];
    e[cnt].to = v;
    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 rt)
{
    lazy[rt<<1] = lazy[rt];
    lazy[rt<<1|1] = lazy[rt];
    tree[rt<<1] = lazy[rt];
    tree[rt<<1|1] = 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(rt);
    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(rt);
    if(l <= mid) res = max(res, query(l, r, lson));
    if(r > mid) res = max(res, query(l, r, rson));
    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(size[v] > maxson) son[u] = v, maxson = size[v];
        }
    }
}
void dfs2(ll u, ll t)
{
    seg[u] = ++num;
    rev[num] = node[u];
    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 == son[u] || v == fa[u]) continue;
        dfs2(v,v);
    }
}
void updRange(ll x, ll y, ll k)
{
    while(top[x] != top[y])
    {
        if(deep[top[x]] < deep[top[y]]) swap(x, y);
        update(seg[top[x]], seg[x], k, 1, n, 1);
        x = fa[top[x]];
    }
    if(deep[x] > deep[y]) swap(x, y);
    update(seg[x], seg[y], k, 1, n, 1);
}
ll qRange(ll x, ll y)
{
    ll ans = 0;
    while(top[x] != top[y])
    {
        if(deep[top[x]] < deep[top[y]]) swap(x, y);
        res = 0;
        res = query(seg[top[x]], seg[x], 1, n, 1);
        ans = max(res, ans);
        x = fa[top[x]];
    }
    if(deep[x] > deep[y]) swap(x, y);
    res = 0;
    res = query(seg[x]+1, seg[y], 1, n, 1);
    ans = max(res, ans);
    return ans;
}
int main()
{
    memset(head, -1, sizeof(head));
    scanf("%lld",&n);
    root = 1;
    for(ll i = 1; i < n; i++)
    {
        scanf("%lld%lld%lld",&u[i],&v[i],&w[i]);
        add(u[i],v[i],w[i]); add(v[i],u[i],w[i]);
    }
    dfs1(root, 0, 1);
    //for(int i = 1; i <= n; i++) cout<<deep[i]<<" ";
    for(ll i = 1; i < n; i++)
    {
        if(deep[u[i]] < deep[v[i]]) node[v[i]] = w[i];
        else node[u[i]] = w[i];
    }
    //for(int i = 1; i <= n; i++) cout<<node[i]<<" ";
    dfs2(root, root);
    build(1, n, 1);
    while(cin>>opt && opt[0] != 'D')
    {
        ll x, y, z;
        if(opt[0] == 'C')
        {
            scanf("%lld%lld",&x,&y);
            if(deep[u[x]] > deep[v[x]]) updRange(u[x], u[x], y);
            else updRange(v[x], v[x], y);
        }
        else
        {
            scanf("%lld%lld",&x,&y);
            if(x == y)
            {
                printf("0\n");
                continue;
            }
            printf("%lld\n",qRange(x,y));
        }
    }
}

【luogu P4114 Qtree1】 题解

标签：res   https   tin   ace   add   swap   mem   std   一个   

原文地址：https://www.cnblogs.com/MisakaAzusa/p/9416364.html