【bzoj3631】[JLOI2014]松鼠的新家

题目大意就是按照他给的顺序遍历树，问每个节点会经过几次，最后的那个要减1

类似前缀和的思想 

从起点x到终点y，只需给x,y两个结点加1，给LCA(x,y),fa[LCA(x,y)]减1，最后做一次从底到根的递推即可求出每个点在多少条链上 

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
 
typedef long long LL;
 
#define N 300010
 
struct edge
{
    int to,next;
}e[N<<2];
int head[N<<2];
int cnt;
 
int n;
int id;
int x,y;
 
int a[N],ans[N];
int siz[N],dep[N],fa[N],top[N],son[N],dfn[N];
 
inline int getint()
{
    int x=0,f=1;char ch=getchar();
    while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
 
void link(int x,int y)
{
    e[++cnt]=(edge){y,head[x]};
    head[x]=cnt;
}
 
void dfs(int x)
{
    siz[x]=1;
    son[x]=0;
    dfn[++id]=x;
    for (int i=head[x];i;i=e[i].next)
    {
        int v=e[i].to;
        if (fa[x]!=v)
        {
            fa[v]=x;
            dep[v]=dep[x]+1;
            dfs(v);
            siz[x]+=siz[v];
            if (siz[v]>siz[son[x]])
                son[x]=v;
        }
    }
}
 
void dfs2(int x,int d)
{
    top[x]=d;
    if (son[x])
        dfs2(son[x],d);
    for (int i=head[x];i;i=e[i].next)
    {
        int v=e[i].to;
        if (fa[x]!=v && son[x]!=v)
            dfs2(v,v);
    }
}
 
int lca(int x,int y)
{
    while (top[x]!=top[y])
    {
        int a=top[x];
        int b=top[y];
        if (dep[a]<dep[b])
            swap(a,b),swap(x,y);
        x=fa[top[x]];
    }
    return dep[x]<dep[y] ? x : y;
}
 
int main()
{
    n=getint();
    for (int i=1;i<=n;i++)
        a[i]=getint();
    for (int i=1;i<n;i++)
    {
        x=getint();
        y=getint();
        link(x,y);
        link(y,x);
    }
    dfs(1);
    dfs2(1,1);
    for (int i=2;i<=n;i++)
    {
        int u=a[i-1],v=a[i],w=lca(u,v);
        ans[u]++;
        ans[v]++;
        ans[w]--;
        ans[fa[w]]--;
    }
    for (int i=n;i>=1;i--)
        ans[fa[dfn[i]]]+=ans[dfn[i]];
    for (int i=1;i<=n;i++)
        if (a[1]!=i)
            printf("%d\n",ans[i]-1);
        else
            printf("%d\n",ans[i]);
    return 0;
}