SPOJ GRASSPLA Grass Planting

最新推荐文章于 2020-04-16 20:41:09 发布

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最新推荐文章于 2020-04-16 20:41:09 发布

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分类专栏：树链剖分文章标签： spoj 树链剖分 acm

本文链接：https://blog.youkuaiyun.com/discreeter/article/details/51429571

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本文介绍了一道关于树链剖分与线段树区间求和的经典算法题，详细展示了如何通过树链剖分和线段树来解决树上的边权更新和查询问题。文章提供了一个完整的C++实现代码示例。

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题目：http://www.spoj.com/problems/GRASSPLA/en/

题意：给定一棵树，树有边权，初始边权都为0。有两种操作：第一种是两点之间的边权均加1，第二种是求两点之间的边权和。

思路：树链剖分啊，线段树区间求和。写成多实例一直莫名RE，改成单实例就过了

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

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 100010;
struct edge
{
    int to, next;
} g[N*2];
struct node
{
    int l, r;
    ll sum, mark;
} s[N*4];
int dep[N], top[N], son[N], siz[N], fat[N], id[N], head[N];
int n, m, cnt, num;
void add_edge(int v, int u)
{
    g[cnt].to = u;
    g[cnt].next = head[v];
    head[v] = cnt++;
}
void dfs1(int v, int fa, int d)
{
    dep[v] = d, siz[v] = 1, fat[v] = fa, son[v] = 0;
    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 push_up(int k)
{
    s[k].sum = s[k<<1].sum + s[k<<1|1].sum;
}
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].sum += (s[k<<1].r - s[k<<1].l + 1) * s[k].mark;
        s[k<<1|1].sum += (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].mark = 0, s[k].sum = 0;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(l, mid, k << 1);
    build(mid + 1, r, k << 1|1);
    push_up(k);
}
void update(int l, int r, int k)
{
    if(l <= s[k].l && s[k].r <= r)
    {
        s[k].sum += s[k].r - s[k].l + 1;
        s[k].mark += 1;
        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);
    push_up(k);
}
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);
}
ll query(int l, int r, int k)
{
    if(l <= s[k].l && s[k].r <= r)
        return s[k].sum;
    push_down(k);
    int mid = (s[k].l + s[k].r) >> 1;
    ll ans = 0;
    if(l <= mid) ans += query(l, r, k << 1);
    if(r > mid) ans += query(l, r, k << 1|1);
    push_up(k);
    return ans;
}
ll seek(int v, int u)
{
    int t1 = top[v], t2 = top[u];
    ll ans = 0;
    while(t1 != t2)
    {
        if(dep[t1] < dep[t2])
            swap(t1, t2), swap(v, u);
        ans += query(id[t1], id[v], 1);
        v = fat[t1], t1 = top[v];
    }
    if(v == u) return ans;
    if(dep[v] > dep[u]) swap(v, u);
    return ans + query(id[son[v]], id[u], 1);
}
void slove()
{
    char ch;
    int a, b;
    while(m--)
    {
        scanf(" %c%d%d", &ch, &a, &b);
        if(ch == 'P') renew(a, b);
        else printf("%lld\n", seek(a, b));
    }
}
int main()
{
    int a, b;
    scanf("%d%d", &n, &m);
    memset(head, -1, sizeof head);
    cnt = num = 0;
    for(int i = 1; i <= n - 1; i++)
    {
        scanf("%d%d", &a, &b);
        add_edge(a, b);
        add_edge(b, a);
    }
    dfs1(1, 0, 1);
    dfs2(1, 1);
    build(1, num, 1);
    slove();

    return 0;
}