[xyz模拟题]动态维护树的直径

专出神题的xyz。

支持删加边、修改点权、维护树的直径。

LCT　　需要额外记录子树信息。用一个堆维护。

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<set>

#include<vector>

#include<queue>

using namespace std;

#define rep(i,x,y) for(i=x;i<=y;i++)

#define _rep(i,x,y) for(i=x;i>=y;i--)

#define REP(i,x,y) for(int i=(x);i<=(y);i++)

#define ALL(x,S) for(x=S.begin();x!=S.end();x++)

#define mp make_pair

#define fi first

#define se second

#define pb push_back

template<class T> inline void read(T&x){bool fu=0;char c;for(c=getchar();c<=32;c=getchar());if(c==‘-‘)fu=1,c=getchar();for(x=0;c>32;c=getchar())x=x*10+c-‘0‘;if(fu)x=-x;};

template<class T> inline void read(T&x,T&y){read(x);read(y);}

template<class T> inline void read(T&x,T&y,T&z){read(x);read(y);read(z);}

inline char getc(){char c;for(c=getchar();c<=32;c=getchar());return c;}

typedef long long ll;

typedef pair<int,int> pii;

int max(int a,int b,int c){return max(max(a,b),c);}
const int N=100010;

const int inf=int(1e9);

int n,m,i,j,k,x,y,u,v,w0,w1;
multiset<int> son[N],AN;

multiset<int>::iterator pos;
priority_queue<pii> Q;
int ch[N][2],fa[N],key[N];bool rev[N];

int lw[N],rw[N],ans[N],len[N];
void Hins(int i,int t){son[i].insert(lw[t]);}

void Hdel(int i,int t){son[i].erase(son[i].find(lw[t]));}

int mson(int i){return son[i].empty()?0:*son[i].rbegin();}

int tson(int i){if(son[i].size()<=1)return mson(i);pos=--son[i].end();return *pos+*(--pos);}
bool tp(int i){return !(ch[fa[i]][0]==i||ch[fa[i]][1]==i);}

void REV(int i){if(i){rev[i]=!rev[i];swap(ch[i][0],ch[i][1]);swap(lw[i],rw[i]);}}

void D(int i){if(rev[i])rev[i]=0,REV(ch[i][0]),REV(ch[i][1]);}

void U(int i)

{

    #define L ch[i][0]

    #define R ch[i][1]

    len[i]=len[L]+key[i]+len[R];

    ans[i]=max(rw[L]+lw[R],max(rw[L],lw[R])+mson(i),tson(i))+key[i];

    lw[i]=max(lw[L],len[L]+key[i]+mson(i),len[L]+key[i]+lw[R]);

    rw[i]=max(rw[R],len[R]+key[i]+mson(i),len[R]+key[i]+rw[L]);

    Q.push(mp(ans[i],i));

}
void rot(int i,int t){int x,y;x=ch[i][!t];D(x);y=ch[x][t];ch[i][!t]=y;ch[x][t]=i;if(y)fa[y]=i;if(fa[i])if(ch[fa[i]][0]==i)ch[fa[i]][0]=x;else if(ch[fa[i]][1]==i)ch[fa[i]][1]=x;fa[x]=fa[i];fa[i]=x;U(i);}

void spl(int x){int y,z;D(x);while(!tp(x)){y=fa[x];z=fa[y];if(z)D(z);if(y)D(y);if(tp(y))rot(y,ch[y][0]==x);else if(ch[z][0]==y)if(ch[y][0]==x)rot(z,1),rot(y,1);else rot(y,0),rot(z,1);else if(ch[y][0]==x)rot(y,1),rot(z,0);else rot(z,0),rot(y,0);U(x);}}

void acc(int x){for(int i=0,t;x;i=x,x=fa[x]){spl(x);t=ch[x][1];if(t)Hins(x,t);ch[x][1]=i;if(i){fa[i]=x;Hdel(x,i);}U(x);}}

void maker(int x){acc(x);spl(x);REV(x);}

void change(int x,int y){acc(x);spl(x);key[x]=y;U(x);}

void cut(int x,int y){maker(x);acc(y);spl(y);ch[y][0]=0;fa[x]=0;U(y);}

void link(int x,int y){maker(x);acc(y);spl(y);fa[x]=y;Hins(y,x);U(y);}

vector<int> E[N];void dfs(int i){vector<int>::iterator it;ALL(it,E[i])if((*it)!=fa[i])fa[*it]=i,dfs(*it),Hins(i,*it);U(i);}

void solve(){pii u;while(1){u=Q.top();if(ans[u.se]==u.fi)break;Q.pop();}printf("%d\n",u.fi);}
int main()

{

    freopen("longest.in","r",stdin);freopen("longest.out","w",stdout);

    read(n,m);

    rep(i,1,n-1)read(x,y),E[x].pb(y),E[y].pb(x);

    rep(i,1,n)read(key[i]);

    dfs(1);

    solve();

    for(;m;m--)

    {

        char c=getc();

        if(c==‘M‘)read(x,y),change(x,y);

        else if(c==‘C‘)read(x,y),read(u,v),cut(x,y),link(u,v);

        solve();

    }

    return 0;

}

[xyz模拟题]动态维护树的直径