树上启发式合并求调
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树上启发式合并求调
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Code_AC楼主2022/8/28 13:43
#include<bits/stdc++.h>
using namespace std;
const int MAXN=1e5+5;

struct edge
{
    int to,nxt;
}e[MAXN<<1];

int head[MAXN],cnt;

inline void add(int x,int y)
{
    e[++cnt].to=y;
    e[cnt].nxt=head[x];
    head[x]=cnt;
    return;
}

int dep[MAXN],siz[MAXN],hson[MAXN];
int n,m;

inline void dfs1(int x,int fa)
{
    dep[x]=dep[fa]+1;
    siz[x]=1;
    int maxson=-1;
    for(register int i=head[x];i;i=e[i].nxt)
    {
        int y=e[i].to;
        if(y==fa)
            continue;
        dfs1(y,x);
        siz[x]+=siz[y];
        if(maxson<siz[y])
        {
            maxson=siz[y];
            hson[x]=y;
        }
    }
    return;
}

int heavy;
int tot[MAXN],ans[MAXN],val[MAXN];
int c[MAXN];

struct Query
{
    int id,y;
};

vector<Query>q[MAXN];

inline void change(int x,int fa,int k)
{
    if(k==-1)
        tot[val[c[x]]]+=k;
    val[c[x]]+=k;
    if(k==1)
        tot[val[c[x]]]+=k;
    for(register int i=head[x];i;i=e[i].nxt)
    {
        int y=e[i].to;
        if(y==fa || y==heavy)
            continue;
        change(y,x,k);
    }
    return;
}

inline void getans(int x)
{
    for(register int i=0;i<q[x].size();i++)
    {
        int id=q[x][i].id;
        int y=q[x][i].y;
        ans[id]=tot[y];
    }
    return;
}

inline void dfs2(int x,int fa,int k)
{
    for(register int i=head[x];i;i=e[i].nxt)
    {
        int y=e[i].to;
        if(y==fa || y==hson[x])
            continue;
        dfs2(y,x,0);
    }
    if(hson[x])
    {
        change(hson[x],x,1);
        heavy=hson[x];
    }
    change(x,fa,1);
    getans(x);
    heavy=0;
    if(k==0)
        change(x,fa,-1);
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    cin>>n>>m;
    for(register int i=1;i<=n;i++)
        cin>>c[i];
    for(register int i=1;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        add(x,y),add(y,x);
    }
    for(register int i=1;i<=m;i++)
    {
        int x,y;
        cin>>x>>y;
        q[x].push_back((Query){i,y});
    }
    dfs1(1,0);
    dfs2(1,0,0);
    for(register int i=1;i<=m;i++)
        printf("%d\n",ans[i]);
    return 0;
}

第一个样例输出

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2022/8/28 13:43
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