New Year Tree CodeForces

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看了题解才懂系列..

利用dfs序将树形结构化为线性结构 先序遍历一棵树得到的序列中每个父节点之后紧跟的都是其子节点 只要知道每个节点有多少子节点 就知道了需要修改的区间的范围

剩下的就是状态压缩了 60种颜色用longlong存即可

#include <bits/stdc++.h>using namespace std;#define ll long longstruct nodeI{    int v;    int next;};struct nodeII{    int l;    int r;    ll val;    ll laz;};nodeI edge[800010];nodeII tree[2000010];ll clr[400010],pre[100];int first[400010],id[400010],mp[400010],sum[400010];int n,q,num;void addedge(int u,int v){    edge[num].v=v;    edge[num].next=first[u];    first[u]=num++;    return;}void dfs(int fa,int u){    int i,v;    id[u]=++num;    for(i=first[u];i!=-1;i=edge[i].next)    {        v=edge[i].v;        if(v!=fa)        {            dfs(u,v);            sum[u]+=(sum[v]+1);        }    }    return;}void pushup(int cur){    tree[cur].val=tree[cur*2].val|tree[cur*2+1].val;    return;}void pushdown(int cur){    if(tree[cur].laz)    {        tree[cur*2].val=tree[cur].laz;        tree[cur*2].laz=tree[cur].laz;        tree[cur*2+1].val=tree[cur].laz;        tree[cur*2+1].laz=tree[cur].laz;        tree[cur].laz=0;    }    return;}void build(int l,int r,int cur){    int m;    tree[cur].l=l;    tree[cur].r=r;    tree[cur].val=0;    tree[cur].laz=0;    if(l==r)    {        tree[cur].val=clr[mp[++num]];        return;    }    m=(l+r)/2;    build(l,m,cur*2);    build(m+1,r,cur*2+1);    pushup(cur);    return;}void update(int pl,int pr,ll val,int cur){    if(pl<=tree[cur].l&&tree[cur].r<=pr)    {        tree[cur].val=val;        tree[cur].laz=val;        return;    }    pushdown(cur);    if(pl<=tree[cur*2].r) update(pl,pr,val,cur*2);    if(pr>=tree[cur*2+1].l) update(pl,pr,val,cur*2+1);    pushup(cur);    return;}ll query(int pl,int pr,int cur){    ll res;    if(pl<=tree[cur].l&&tree[cur].r<=pr)    {        return tree[cur].val;    }    pushdown(cur);    res=0;    if(pl<=tree[cur*2].r) res|=query(pl,pr,cur*2);    if(pr>=tree[cur*2+1].l) res|=query(pl,pr,cur*2+1);    pushup(cur);    return res;}int judge(ll p){    int res;    res=0;    while(p>0)    {        if(p%2==1) res++;        p/=2;    }    return res;}void init(){    int i;    pre[0]=1;    for(i=1;i<=60;i++)    {        pre[i]=pre[i-1]*2;    }    return;}int main(){    ll tem;    int i,op,u,v,c;    init();    while(scanf("%d%d",&n,&q)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d",&c);            clr[i]=pre[c];        }        memset(first,-1,sizeof(first));        num=0;        for(i=1;i<=n-1;i++)        {            scanf("%d%d",&u,&v);            addedge(u,v);            addedge(v,u);        }        memset(sum,0,sizeof(sum));        num=0;        dfs(-1,1);        for(i=1;i<=n;i++)        {            mp[id[i]]=i;        }        num=0;        build(1,n,1);        while(q--)        {            scanf("%d",&op);            if(op==1)            {                scanf("%d%d",&u,&c);                update(id[u],id[u]+sum[u],pre[c],1);            }            else            {                scanf("%d",&u);                tem=query(id[u],id[u]+sum[u],1);                printf("%d\n",judge(tem));            }        }    }    return 0;}