HDU - 3966-Aragorn' Story(树链剖分+线段树)-优快云博客

本文详细介绍了使用树链剖分解决复杂数据更新和查询问题的方法。通过实例解析，展示了如何在敌军动态调整兵力的情况下，实时计算特定营地的士兵数量。文章提供了完整的代码实现，包括树链剖分的初始化、DFS遍历、线段树构建与更新等关键步骤。

链接：https://vjudge.net/problem/HDU-3966

题意：

Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time.

思路：

第一次写树链剖分，直接上模板。

代码：

#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
#include <set>
#include <iterator>
#include <cstring>
#include <assert.h>
using namespace std;

typedef long long LL;
const int MAXN = 5e4+10;
vector<int> G[MAXN];
int Dis[MAXN], Fa[MAXN], Top[MAXN], Size[MAXN];
int Son[MAXN], Id[MAXN], Rk[MAXN];
int Seg[MAXN*4], A[MAXN], Add[MAXN*4];
int n, m, p;
int x, y;
int cnt = 0;

void Init()
{
    for (int i = 1;i <= n;i++)
        G[i].clear();
    memset(Seg, 0, sizeof(Seg));
    memset(Son, 0, sizeof(Son));
    memset(Add, 0, sizeof(Add));
    cnt = 0;
}

void Dfs1(int x, int u, int dep)
{
    Dis[u] = dep;
    Fa[u] = x;
    Size[u] = 1;
    for (int i = 0;i < G[u].size();i++)
    {
        int v = G[u][i];
        if (v == x)
            continue;
        Dfs1(u, v, dep+1);
        Size[u] += Size[v];
        if (Size[v] > Size[Son[u]])
            Son[u] = v;
    }
}

void Dfs2(int u, int top)
{
    Top[u] = top;
    Id[u] = ++cnt;
    Rk[cnt] = u;
    if (!Son[u])
        return;
    Dfs2(Son[u], top);
    for (int i = 0;i < G[u].size();i++)
    {
        int v = G[u][i];
        if (v != Son[u] && v != Fa[u])
            Dfs2(v, v);
    }
}

void PushUp(int root)
{
    Seg[root] = Seg[root<<1]+Seg[root<<1|1];
}

void PushDown(int root, int l, int r)
{
    if (Add[root])
    {
        int mid = (l+r)/2;
        Add[root<<1] += Add[root];
        Add[root<<1|1] += Add[root];
        Seg[root<<1] += (mid-l+1)*Add[root];
        Seg[root<<1|1] += (r-mid)*Add[root];
        Add[root] = 0;
    }
}

void Build(int root, int l, int r)
{
    if (l == r)
    {
        Seg[root] = A[Rk[l]];
        return;
    }
    int mid = (l+r)/2;
    Build(root<<1, l, mid);
    Build(root<<1|1, mid+1, r);
    PushUp(root);
}

int Query(int root, int l, int r, int ql, int qr)
{
    if (r < ql || qr < l)
        return 0;
    if (ql <= l && r <= qr)
        return Seg[root];
    PushDown(root, l, r);
    int mid = (l+r)/2;
    int res = 0;
    res += Query(root<<1, l, mid, ql, qr);
    res += Query(root<<1|1, mid+1, r, ql, qr);
    PushUp(root);
    return res;
}

void Update(int root, int l, int r, int ql, int qr, int c)
{
    if (r < ql || qr < l)
        return;
    if (ql <= l && r <= qr)
    {
        Seg[root] += (r-l+1)*c;
        Add[root] += c;
        return;
    }
    PushDown(root, l, r);
    int mid = (l+r)/2;
    Update(root<<1, l, mid, ql, qr, c);
    Update(root<<1|1, mid+1, r, ql, qr, c);
    PushUp(root);
}

void UpdateLine(int l, int r, int c)
{
    while (Top[l] != Top[r])
    {
        if (Dis[Top[l]] < Dis[Top[r]])
            swap(l, r);
        Update(1, 1, n, Id[Top[l]], Id[l], c);
        l = Fa[Top[l]];
    }
    if (Id[l] < Id[r])
        Update(1, 1, n, Id[l], Id[r], c);
    else
        Update(1, 1, n, Id[r], Id[l], c);
}

int main()
{
//    freopen("test.in", "r", stdin);
    while (~scanf("%d%d%d", &n, &m, &p))
    {
        Init();
        for (int i = 1;i <= n;i++)
            scanf("%d", &A[i]);
        for (int i = 1;i < n;i++)
        {
            scanf("%d%d", &x, &y);
            G[x].push_back(y);
            G[y].push_back(x);
        }
        Dfs1(0, 1, 1);
        Dfs2(1, 1);
        Build(1, 1, n);
//        for (int i = 1;i <= n;i++)
//            cout << Dis[i] << ' ';
//        cout << endl;
//        for (int i = 1;i <= n;i++)
//            cout << Id[i] << ' ';
//        cout << endl;
//        for (int i = 1;i <= n;i++)
//            cout << Top[i] << ' ' ;
//        cout << endl;
        char op[10];
        int l, r, v, w;
        while (p--)
        {
            scanf("%s", op);
            if (op[0] == 'I')
            {
                scanf("%d%d%d", &l, &r, &v);
//                cout << Top[l] << ' ' << Top[r] << endl;
                UpdateLine(l, r, v);
            }
            else if (op[0] == 'D')
            {
                scanf("%d%d%d", &l, &r, &v);
                v = -v;
                UpdateLine(l, r, v);
            }
            else
            {
                scanf("%d", &w);
                printf("%d\n", Query(1, 1, n, Id[w], Id[w]));
            }
        }
    }

    return 0;
}

转载于:https://www.cnblogs.com/YDDDD/p/10971403.html