[ZJOI2008]树的统计Count

题目大意:

给出一棵树,点上有点权

有两种询问,询问两点之间树链上的点权和,询问两点之间树链上最大点权

点数<=30000,询问树<=200000


最裸的动态树,操作少,维护的值简单


//Lib
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<ctime>
 
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
using namespace std;
//Macro
#define rep(i,a,b) for(int i=a,tt=b;i<=tt;++i)
#define rrep(i,a,b) for(int i=a,tt=b;i>=tt;--i)
#define erep(i,e,x) for(int i=x;i;i=e[i].next)
#define irep(i,x) for(__typedef(x.begin()) i=x.begin();i!=x.end();i++)
#define read() (strtol(ipos,&ipos,10))
#define sqr(x) ((x)*(x))
#define pb push_back
#define PS system("pause");
typedef long long ll;
typedef pair<int,int> pii;
const int oo=100000000;
const double inf=1e20;
const double eps=1e-6;
string name="",in=".in",out=".out";
//Var
struct E
{
    int next,node;
}e[60008];
struct T
{
    int LC,RC,KEY,SUM,MAX,FA;
    #define lc(x) tree[x].LC
    #define rc(x) tree[x].RC
    #define key(x) tree[x].KEY
    #define sum(x) tree[x].SUM
    #define Max(x) tree[x].MAX
    #define fa(x) tree[x].FA
}tree[30008];
int tot,n,Q,w[30008],h[30008];
bool vis[30008];
void add(int a,int b){e[++tot].next=h[a];e[tot].node=b;h[a]=tot;}
bool isRoot(int x){return (lc(fa(x))!=x)&&(rc(fa(x))!=x);}
void Update(int x)
{
    sum(x)=sum(lc(x))+sum(rc(x))+key(x);
    Max(x)=max(max(Max(lc(x)),Max(rc(x))),key(x));
}
void Zig(int x)
{
    int y=fa(x),z=fa(y);
    if(lc(z)==y)lc(z)=x;else if(rc(z)==y)rc(z)=x;
    fa(x)=z;fa(rc(x))=y;lc(y)=rc(x);rc(x)=y;fa(y)=x;
    Update(y);
}
void Zag(int x)
{
    int y=fa(x),z=fa(y);
    if(lc(z)==y)lc(z)=x;else if(rc(z)==y) rc(z)=x;
    fa(x)=z;fa(lc(x))=y;rc(y)=lc(x);lc(x)=y;fa(y)=x;
    Update(y);
}
void Splay(int x) {
    int y , z;
    while (!isRoot(x)) {
        y = fa(x); z = fa(y);
        if (isRoot(y))
            if (x == lc(y)) Zig(x);
            else Zag(x);
        else
            if (y == lc(z))
                if (x == lc(y)) Zig(y) , Zig(x);
                else Zag(x) , Zig(x);
            else
                if (x == lc(y)) Zig(x) , Zag(x);
                else Zag(y) , Zag(x);
    }
    Update(x);
}
void DFS(int u,int f)
{
    key(u)=w[u];
    erep(i,e,h[u])if(e[i].node!=f)
        DFS(e[i].node,u),fa(e[i].node)=u;
}
void Init()
{
    int a,b;
    scanf("%d",&n);
    rep(i,1,n-1)
    {
        scanf("%d%d",&a,&b);
        add(a,b);add(b,a);
    }
    rep(i,1,n)scanf("%d",w+i);
    key(0)=Max(0)=-oo;
    DFS(1,0);
}
void Expose(int x)
{
    int y;
    for(y=0;x;x=fa(x)){Splay(x);rc(x)=y;Update(x);y=x;}
}
void Change(int x,int y)
{
    key(x)=y;Splay(x);
}
int Qmax(int x,int y)
{
    Expose(y);
    for(y=0;x;x=fa(x))
    {
        Splay(x);
        if(!fa(x))return max(key(x),max(Max(rc(x)),Max(y)));
        rc(x)=y;Update(x);y=x;
    }
}
int Qsum(int x,int y)
{
    Expose(y);
    for(y=0;x;x=fa(x))
    {
        Splay(x);
        if(!fa(x))return sum(rc(x))+key(x)+sum(y);
        rc(x)=y;Update(x);y=x;
    }
}
void Work()
{
    char ch[10];int a,b;
    scanf("%d",&Q);
    rep(i,1,Q)
    {
        scanf("%s%d%d",ch,&a,&b);
        if(ch[0]=='C')Change(a,b);
        else
        {
            if(ch[1]=='M')printf("%d\n",Qmax(a,b));
            else printf("%d\n",Qsum(a,b));
        }
    }
}
int main()
{
    //freopen((name+in).c_str(),"r",stdin);
    //freopen((name+out).c_str(),"w",stdout);
    Init();
    Work();
    return 0;
}


你可能感兴趣的:(struct,String,tree,OO,BI,System)