BZOJ3631: [JLOI2014]松鼠的新家

传送门

树上的差分优化，很简单的一道题，应该属于NOIP2015TGD2T3的子问题。

//BZOJ 3631
//by Cydiater
//2016.10.25
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
#include <bitset>
#include <iomanip>
using namespace std;
#define ll long long
#define up(i,j,n)		for(int i=j;i<=n;i++)
#define down(i,j,n)		for(int i=j;i>=n;i--)
const int MAXN=6e5+5;
const int oo=0x3f3f3f3f;
inline int read(){
	char ch=getchar();int x=0,f=1;
	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;
}
int N,a[MAXN],LINK[MAXN],lable[MAXN],len=0,fa[MAXN][25],dep[MAXN];
struct edge{
	int y,next;
}e[MAXN];
namespace solution{
	inline void insert(int x,int y){e[++len].next=LINK[x];LINK[x]=len;e[len].y=y;}
	void init(){
		N=read();
		up(i,1,N)a[i]=read();
		up(i,1,N-1){
			int x=read(),y=read();
			insert(x,y);
			insert(y,x);
		}
	}
	void dfs(int node,int deep,int father){
		fa[node][0]=father;dep[node]=deep;
		for(int i=LINK[node];i;i=e[i].next)if(e[i].y!=father)
			dfs(e[i].y,deep+1,node);
	}
	void get_ancestor(){
		up(i,1,20)up(node,1,N)if(fa[node][i-1]!=0)
			fa[node][i]=fa[fa[node][i-1]][i-1];
	}
	int LCA(int x,int y){
		if(x==y)		return x;
		if(dep[x]<dep[y])swap(x,y);
		down(i,20,0)if(dep[x]-(1<<i)>=dep[y])x=fa[x][i];
		if(x==y)		return x;
		down(i,20,0)if(fa[x][i]!=0&&fa[x][i]!=fa[y][i]){
			x=fa[x][i];y=fa[y][i];
		}
		return fa[x][0];
	}
	void re_dfs(int node){
		for(int i=LINK[node];i;i=e[i].next)if(e[i].y!=fa[node][0]){
			re_dfs(e[i].y);
			lable[node]+=lable[e[i].y];
		}
	}
	void slove(){
		dfs(1,0,0);
		get_ancestor();
		up(i,1,N-1){
			int x=a[i],y=a[i+1],lca=LCA(x,y);
			lable[x]++;lable[y]++;lable[fa[lca][0]]--;lable[lca]--;
		}
		re_dfs(1);
		up(i,2,N)lable[a[i]]--;
	}
	void output(){
		up(i,1,N)printf("%d\n",lable[i]);
	}
}
int main(){
	//freopen("input.in","r",stdin);
	using namespace solution;
	init();
	slove();
	output();
	return 0;
}