CodeForces 191C Fools and Roads 树链剖分

最新推荐文章于 2021-11-07 16:19:08 发布

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最新推荐文章于 2021-11-07 16:19:08 发布

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分类专栏：树链剖分文章标签： acm Codeforces 树链剖分线段树

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本文介绍了一种解决树形结构上路径更新问题的方法——树链剖分算法。通过预处理将树分解成一系列链，并利用线段树进行区间更新与查询，实现高效的边权更新。适用于Codeforces 191/C题目。

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题目:http://codeforces.com/problemset/problem/191/C

题意：给定一棵树，有n个节点n - 1条边，然后每次给出两个节点，两点路径上的边的权加1（初始边权为0），最后依次输出每一条边的边权

思路：树链剖分，每次更新时lazy操作，最后依次查询边权就好了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 100010;
struct edge
{
    int to, next;
}g[N*2];
struct node
{
    int l, r, val, mark;
}s[N*4];
int dep[N], top[N], son[N], siz[N], fat[N], id[N], head[N];
int d[N][3], val[N];
int n, k, cnt, num;
void dfs1(int v, int fa, int d)
{
    dep[v] = d, siz[v] = 1, son[v] = 0, fat[v] = fa;
    for(int i = head[v]; i != -1; i = g[i].next)
    {
        int u = g[i].to;
        if(u != fa)
        {
            dfs1(u, v, d + 1);
            siz[v] += siz[u];
            if(siz[son[v]] < siz[u]) son[v] = u;
        }
    }
}
void dfs2(int v, int tp)
{
    top[v] = tp, id[v] = ++num;
    if(son[v]) dfs2(son[v], tp);
    for(int i = head[v]; i != -1; i = g[i].next)
    {
        int u = g[i].to;
        if(u != fat[v] && u != son[v]) dfs2(u, u);
    }
}
void add_edge(int v, int u)
{
    g[cnt].to = u;
    g[cnt].next = head[v];
    head[v] = cnt++;
}
void push_down(int k)
{
    if(s[k].mark)
    {
        s[k<<1].mark += s[k].mark;
        s[k<<1|1].mark += s[k].mark;
        s[k<<1].val += (s[k<<1].r - s[k<<1].l + 1) * s[k].mark;
        s[k<<1|1].val += (s[k<<1|1].r - s[k<<1|1].l + 1) * s[k].mark;
        s[k].mark = 0;
    }
}
void build(int l, int r, int k)
{
    s[k].l = l, s[k].r = r, s[k].val = 0, s[k].mark = 0;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(l, mid, k << 1);
    build(mid + 1, r, k << 1|1);
}
void update(int l, int r, int k)
{
    if(l <= s[k].l && s[k].r <= r)
    {
        s[k].val += s[k].r - s[k].l + 1;
        s[k].mark++;
        return;
    }
    push_down(k);
    int mid = (s[k].l + s[k].r) >> 1;
    if(l <= mid) update(l, r, k << 1);
    if(r > mid) update(l, r, k << 1|1);
}
void renew(int v, int u)
{
    int t1 = top[v], t2 = top[u];
    while(t1 != t2)
    {
        if(dep[t1] < dep[t2])
            swap(t1, t2), swap(v, u);
        update(id[t1], id[v], 1);
        v = fat[t1], t1 = top[v];
    }
    if(v == u) return;
    if(dep[v] > dep[u]) swap(v, u);
    update(id[son[v]], id[u], 1);
}
int query(int v, int k)
{
    if(s[k].l == s[k].r)
        return s[k].val;
    push_down(k);
    int mid = (s[k].l + s[k].r) >> 1;
    if(v <= mid) return query(v, k << 1);
    else return query(v, k << 1|1);
}
void slove()
{
    int a, b;
    build(1, num, 1);
    scanf("%d", &k);
    for(int i = 0; i < k; i++)
    {
        scanf("%d%d", &a, &b);
        renew(a, b);
    }
    for(int i = 1; i <= n - 1; i++)
    {
        int res = query(id[d[i][1]], 1);
        if(i == 1) printf("%d", res);
        else printf(" %d", res);
    }
}
int main()
{
    scanf("%d", &n);
    memset(head, -1, sizeof head);
    cnt = num = 0;
    for(int i = 1; i <= n - 1; i++)
    {
        scanf("%d%d", &d[i][0], &d[i][1]);
        add_edge(d[i][0], d[i][1]);
        add_edge(d[i][1], d[i][0]);
    }
    dfs1(1, 0, 1);
    dfs2(1, 1);
    for(int i = 1; i <= n; i++)
        if(dep[d[i][0]] > dep[d[i][1]]) swap(d[i][0], d[i][1]);
    slove();

    return 0;
}