Codeforces 620E New Year Tree (哈希 + 线段树)

http://codeforces.com/problemset/problem/620/E

                       New Year Tree

The New Year holidays are over, but Resha doesn’t want to throw away the New Year tree. He invited his best friends Kerim and Gural to help him to redecorate the New Year tree.

The New Year tree is an undirected tree with n vertices and root in the vertex 1.

You should process the queries of the two types:

  1. Change the colours of all vertices in the subtree of the vertex v to the colour c.
  2. Find the number of different colours in the subtree of the vertex v.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 4·105) — the number of vertices in the tree and the number of the queries.

The second line contains n integers ci (1 ≤ ci ≤ 60) — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the vertices of the j-th edge. It is guaranteed that you are given correct undirected tree.

The last m lines contains the description of the queries. Each description starts with the integer tk (1 ≤ tk ≤ 2) — the type of the k-th query. For the queries of the first type then follows two integers vk, ck (1 ≤ vk ≤ n, 1 ≤ ck ≤ 60) — the number of the vertex whose subtree will be recoloured with the colour ck. For the queries of the second type then follows integer vk (1 ≤ vk ≤ n) — the number of the vertex for which subtree you should find the number of different colours.

Output

For each query of the second type print the integer a — the number of different colours in the subtree of the vertex given in the query.

Each of the numbers should be printed on a separate line in order of query appearing in the input.
Sample test(s)
Input

7 10
1 1 1 1 1 1 1
1 2
1 3
1 4
3 5
3 6
3 7
1 3 2
2 1
1 4 3
2 1
1 2 5
2 1
1 6 4
2 1
2 2
2 3

Output

2
3
4
5
1
2


一棵无向树(以1为根)。放对一个节点染色的时候,该节点的子树都被染成这个颜色。多次询问,每个询问为输入该树的某个节点然后输出它和它的子树有多少种颜色。

先将树dfs一次,把节点用dfs序表达出来。哈希成一段连续的序列。记住每个节点的区间范围。

我写这篇博客的目的是 因为这个地方WA n次:
由于颜色有60种,用二进制的 1<<60表示。
但是这样写 1 << col会WA
正确的姿势为: 1LL << col …..


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

#define lson l,mid,i<<1
#define rson mid+1,r,i<<1|1

const int N = 4 * 1e5 + 500;

struct Node
{
    int next,to;
    Node(){}
    Node(int a,int b):next(a),to(b){}
};
struct Node edge[N*2]; int head[N];  int nedges = -1;
void add(int a,int b)
{
    edge[++nedges] = Node(head[a],b);
    head[a] = nedges;
    edge[++nedges] = Node(head[b],a);
    head[b] = nedges;
}

bool vis[N];  int has[N]; int a[N]; int in[N];  int out[N]; int mark = 1;

void dfs(int u)
{
    in[u] = mark; has[mark] = u; vis[u] = true;
    for(int i=head[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].to;
        if( vis[v] ) continue;
        mark++; dfs(v);
    }
    out[u] = mark;
}
long long ans;
struct tree 
{
    long long val;
    int flag;
}tr[N*4];

void built(int l,int r,int i)
{
    if(l==r)
    {
        tr[i].val = 1LL<<(a[has[l]]);  // !!!!!
        tr[i].flag=0;
        return;
    }

    int mid=(l+r)>>1;
    built(lson);
    built(rson);
    tr[i].val = tr[i<<1].val | tr[i<<1|1].val ;
    return;
}
void pushdown(int i,int col)
{
    tr[i<<1].flag=tr[i<<1|1].flag=col;
    tr[i<<1].val=tr[i<<1|1].val=1LL<<col;
    tr[i].flag=0;
    return;
}
void update(int l,int r,int i,int a,int b,int col)
{
    if(l>=a&&r<=b)
    {
        tr[i].val = 1LL<<col;
        tr[i].flag=col;
        return;
    }
    int mid=(l+r)>>1;
    if(tr[i].flag) pushdown(i,tr[i].flag);
    if(mid>=a) update(lson,a,b,col);
    if(mid<b) update(rson,a,b,col);
    tr[i].val = tr[i<<1].val | tr[i<<1|1].val;
    return;
}
void query(int l,int r,int i,int a,int b)
{
    if(l>=a&&r<=b)
    {
        ans|= tr[i].val;
        return;
    }

    int mid=(l+r)>>1;
    if(tr[i].flag) pushdown(i,tr[i].flag);
    if(mid>=a) query(lson,a,b);
    if(mid<b) query(rson,a,b);
    return;
}

int main()
{
    int n,m; scanf("%d%d",&n,&m);
    memset(vis,false,sizeof(vis));
    memset(head,-1,sizeof(head));
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<n;i++)
    {
        int u,v; scanf("%d%d",&u,&v);
        add(u,v);
    }
    dfs(1);
    built(1,n,1);
    while(m--)
    {
       int op;  scanf("%d",&op);
       if(op==1){
           int x,col; scanf("%d%d",&x,&col);
           update(1,n,1,in[x],out[x],col);
       }
       else{
           int x;  scanf("%d",&x);
           ans = 0;
           query(1,n,1,in[x],out[x]);
           int sum = 0;
           for(int i=1;i<=60;i++)
               if( (ans>>i) & 1) sum++;
               printf("%d\n",sum);
       }
    }
    return 0;
}
当前提供的引用内容并未提及关于Codeforces比赛M1的具体时间安排[^1]。然而,通常情况下,Codeforces的比赛时间会在其官方网站上提前公布,并提供基于不同时区的转换工具以便参赛者了解具体开赛时刻。 对于Codeforces上的赛事而言,如果一场名为M1的比赛被计划举行,则它的原始时间一般按照UTC(协调世界时)设定。为了得知该场比赛在UTC+8时区的确切开始时间,可以遵循以下逻辑: - 前往Codeforces官网并定位至对应比赛页面。 - 查看比赛所标注的标准UTC起始时间。 - 将此标准时间加上8小时来获取对应的北京时间(即UTC+8)。 由于目前缺乏具体的官方公告链接或者确切日期作为依据,无法直接给出Codeforces M1比赛于UTC+8下的实际发生时段。建议定期访问Codeforces平台查看最新动态更新以及确认最终版程表信息。 ```python from datetime import timedelta, datetime def convert_utc_to_bj(utc_time_str): utc_format = "%Y-%m-%dT%H:%M:%SZ" bj_offset = timedelta(hours=8) try: # 解析UTC时间为datetime对象 utc_datetime = datetime.strptime(utc_time_str, utc_format) # 转换为北京时区时间 beijing_time = utc_datetime + bj_offset return beijing_time.strftime("%Y-%m-%d %H:%M:%S") except ValueError as e: return f"错误:{e}" # 示例输入假设某场Codeforces比赛定于特定UTC时间 example_utc_start = "2024-12-05T17:35:00Z" converted_time = convert_utc_to_bj(example_utc_start) print(f"Codeforces比赛在北京时间下将是:{converted_time}") ```
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