New Year Tree CodeForces

来源：互联网发布：淘宝发布宝贝教程2016 编辑：程序博客网时间：2024/06/06 01:54

题意：

给出区间内数字，有两个操作。操作一更改X的所有子树，操作二询问子树中不同数字个数。

思路：

之前没想到二进制优化，本来认为60个气球很少，就开个数组维护了个vis标记，然而TLE了。上网查说的很有道理60 完全可以用2^60的long long 存在，这样就很好做了！

#include <iostream>#include <stdio.h>#include <map>#include <cstring>#include <queue>#include <vector>using namespace std;const int maxn=500005;vector<int> vec[maxn];int vis[maxn];int id[maxn];int post[maxn];int a[maxn];int topw=0;long long ans;struct node{    int left,right;    long long num,lazy;}tree[maxn*4];void dfs(int u){    vis[u]=1;    id[u]=++topw;    for(int i=0;i<vec[u].size();i++)    {        int v=vec[u][i];        if(vis[v]) continue;        dfs(v);    }    post[u]=topw;}void push_down(int i){    if(!tree[i].lazy)        return ;    tree[i << 1].num = tree[i << 1 | 1].num = (1LL << tree[i].lazy);    tree[i << 1].lazy = tree[i << 1 | 1].lazy = tree[i].lazy;    tree[i].lazy = 0;}void push_up(int i){    tree[i].num = tree[i << 1].num | tree[i << 1 | 1].num;}void build(int i,int left,int right){    tree[i].lazy=0;    tree[i].left=left;    tree[i].right=right;    if(left==right)    {        return ;    }    int mid=(tree[i].left+tree[i].right)>>1;    build(i<<1,left,mid);    build(i<<1|1,mid+1,right);}void get_pos(int i,int aim,int w){    if(tree[i].left==aim&&tree[i].right==aim)    {        tree[i].num = (1LL << w );        return ;    }    push_down(i);    int mid=(tree[i].left+tree[i].right)>>1;    if(aim<=mid)        get_pos(i<<1,aim,w);    if(aim>mid)        get_pos(i<<1|1,aim,w );    push_up(i);}void update(int i,int left,int right,long long  w){    if(tree[i].left==left&&tree[i].right==right)    {        tree[i].lazy=w;        tree[i].num=(1LL<<w);        return ;    }    push_down(i);    int mid=(tree[i].left+tree[i].right)>>1;    if(right<=mid)        update(i<<1,left,right,w);    else if(left>mid)        update(i<<1|1,left,right,w);    else    {        update(i<<1,left,mid,w);        update(i<<1|1,mid+1,right,w);    }    push_up(i);    return ;}void query(int i,int left,int right){    if(tree[i].left==left&&tree[i].right==right )    {        ans |= tree[i].num;        return ;    }    push_down(i);    int mid=(tree[i].left+tree[i].right)>>1;    if(right<=mid)    {        query(i<<1,left,right);    }    else if(left>mid)    {        query(i<<1|1,left,right);    }    else    {        query(i<<1,left,mid);        query(i<<1|1,mid+1,right);    }    push_up(i);}int main(){    int n,m;    scanf("%d%d",&n,&m);    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);        vec[u].push_back(v);        vec[v].push_back(u);    }    topw=0;    dfs(1);    build(1,1,topw);    for(int i=1;i<=n;i++)    {        get_pos(1,id[i],a[i]);    }    while(m--)    {        int op;        int x,y;        scanf("%d",&op);        if(op==1)        {            scanf("%d%d",&x,&y);            update(1,id[x],post[x],y);        }        else        {            ans=0;            scanf("%d",&x);            query(1,id[ x ],post[ x ]);            int cnt=0;            for(int i=1;i<=61;i++) {                if(ans & (1LL<<i)) cnt++;            }            printf("%d\n",cnt);        }    }}

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