hdu3974 Assign the task

Assign the task

HDU - 3974

题意：

一棵树

两种操作

1. "C x" 查询节点x的颜色

2."T x y" 将以x为根节点的子树的颜色全部染为k

题解：

dfs序+线段树

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

const int N = 5e5 + 10;
struct node
{
    int l, r;
    int v;
    int lazy;
} tr[N << 2];
int n;

int h[N], ne[N << 1], e[N << 1], idx;
int L[N], R[N], tot;
int rt;
int fa[N];
void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void dfs(int u, int fa)
{
    L[u] = ++tot;
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (j == fa)
            continue;
        dfs(j, u);
    }
    R[u] = tot;
}

void pushdown(int u)
{
    if (tr[u].l == tr[u].r)
    {
        tr[u].lazy = -1;
        return;
    }
    if (tr[u].lazy == 0)
        return;
    tr[u << 1].lazy = 1;
    tr[u << 1 | 1].lazy = 1;
    tr[u << 1].v = tr[u].v;
    tr[u << 1 | 1].v = tr[u].v;
    tr[u].lazy = 0;
}
void build(int u, int l, int r)
{
    if (l == r)
    {
        tr[u] = {l, r, -1, 0};
        return;
    }
    else
    {
        tr[u] = {l, r, -1, 0};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        return;
    }
}

void modify(int u, int l, int r, int x)
{
    if (tr[u].r <= r && tr[u].l >= l)
    {
        tr[u].v = x;
        tr[u].lazy = 1;
        return;
    }
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid)
        modify(u << 1, l, r, x);
    if (r > mid)
        modify(u << 1 | 1, l, r, x);
    return;
}
int query(int u, int x)
{
    // if (x >= tr[u].l && x <= tr[u].r && tr[u].lazy  == 1) return tr[u].v;
    if (tr[u].l == x && tr[u].r == x)
        return tr[u].v;
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (x <= mid)
        return query(u << 1, x);
    else
        return query(u << 1 | 1, x);
}
int main()
{
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    int t;
    cin >> t;
    for (int _ = 1; _ <= t; _++)
    {
        printf("Case #%d:
", _);
        idx = 0;
        tot = 0;
        memset(h, -1, sizeof h);
        memset(fa, 0, sizeof fa);
        cin >> n;
        build(1, 1, n);
        for (int i = 1; i < n; i++)
        {
            int a, b;
            cin >> a >> b;
            add(b, a);
            fa[a] = b;
        }
        for (int i = 1; i <= n; i++)
        {
            if (!fa[i])
            {
                dfs(i, 0);
            }
        }
        // cout << "rt = " << rt  << endl;
        // dfs(rt, 0);
        int m;
        cin >> m;
        while (m--)
        {
            char op[2];
            int x;
            scanf("%s %d", op, &x);
            cin >> op >> x;
            if (op[0] == 'C')
                printf("%d
", query(1, L[x]));
            else
            {
                int k;
                scanf("%d", &k);
                modify(1, L[x], R[x], k);
            }
        }
    }
}