codeforces 620E. New Year Tree dfs序+线段树+bitset

题目链接

给一棵树, 每个节点有颜色, 两种操作, 一种是将一个节点的子树全都染色成c, 一种是查询一个节点的子树有多少个不同的颜色, c<=60.

每个节点一个bitset维护就可以。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 4e5+5;
struct node
{
    int to, nextt;
}e[maxn*2];
int lazy[maxn<<2], a[maxn], in[maxn], out[maxn], head[maxn*2], num, clock;
bitset <63> s[maxn<<2];
void add(int u, int v) {
    e[num].to = v, e[num].nextt = head[u], head[u] = num++;
}
void pushUp(int rt) {
    s[rt] = s[rt<<1]|s[rt<<1|1];
}
void pushDown(int rt) {
    if(lazy[rt]) {
        lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];
        s[rt<<1].reset();
        s[rt<<1|1].reset();
        s[rt<<1][lazy[rt]] = s[rt<<1|1][lazy[rt]] = 1;
        lazy[rt] = 0;
        return ;
    }
}
void update(int val, int L, int R, int l, int r, int rt) {
    if(L<=l&&R>=r) {
        s[rt].reset();
        s[rt][val] = 1;
        lazy[rt] = val;
        return ;
    }
    pushDown(rt);
    int m = l+r>>1;
    if(L<=m)
        update(val, L, R, lson);
    if(R>m)
        update(val, L, R, rson);
    pushUp(rt);
}
void dfs(int u, int fa) {
    in[u] = ++clock;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(v == fa)
            continue;
        dfs(v, u);
    }
    out[u] = clock;
}
bitset<63> query(int L, int R, int l, int r, int rt) {
    if(L<=l&&R>=r) {
        return s[rt];
    }
    pushDown(rt);
    bitset<63> tmp;
    tmp.reset();
    int m = l+r>>1;
    if(L<=m)
        tmp |= query(L, R, lson);
    if(R>m)
        tmp |= query(L, R, rson);
    return tmp;
}
int main()
{
    int q, n, m, x, y;
    int ch;
    memset(head, -1, sizeof(head));
    cin>>n>>m;
    for(int i = 1; i<=n; i++)
        scanf("%d", &a[i]);
    for(int i = 1; i<n; i++) {
        scanf("%d%d", &x, &y);
        add(x, y);
        add(y, x);
    }
    dfs(1, 0);
    for(int i = 1; i<=n; i++) {
        update(a[i], in[i], in[i], 1, n, 1);
    }
    while(m--) {
        scanf("%d%d", &ch, &x);
        if(ch == 2)
            printf("%d
", query(in[x], out[x], 1, n, 1).count());
        else {
            int z;
            scanf("%d", &z);
            update(z, in[x], out[x], 1, n, 1);
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/yohaha/p/5150019.html