SPOJ QTREE4 SPOJ Query on a tree IV

You are given a tree (an acyclic undirected connected graph) with N nodes, and nodes numbered 1,2,3...,N. Each edge has an integer value assigned to it(note that the value can be negative). Each node has a color, white or black. We define dist(a, b) as the sum of the value of the edges on the path from node a to node b.

All the nodes are white initially.

We will ask you to perfrom some instructions of the following form:

• C a : change the color of node a.(from black to white or from white to black)
• A : ask for the maximum dist(a, b), both of node a and node b must be white(a can be equal to b). Obviously, as long as there is a white node, the result will alway be non negative.

Input

• In the first line there is an integer N (N <= 100000)
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of value c (-1000 <= c <= 1000)
• In the next line, there is an integer Q denotes the number of instructions (Q <= 100000)
• In the next Q lines, each line contains an instruction "C a" or "A"

Output

For each "A" operation, write one integer representing its result. If there is no white node in the tree, you should write "They have disappeared.".

Example

Input:
3
1 2 1
1 3 1
7
A
C 1
A
C 2
A
C 3
A

Output:
2
2
0
They have disappeared.

Some new test data cases were added on Apr.29.2008, all the solutions have been rejudged.

树 Link cut tree

查动态点分治时候在别的博客看到一个LCT做Qtree的专题，心生敬仰于是学（chao）了一发。

结论是相比之下还是点分治好写啊，这个状态更新神烦，如果不标准代码比对，我估计得调几个小时吧……

--------

大致就是，把边权迁移到子结点上，先建好LCT。Splay树是二叉树，那么就把当前未激活的边全都扔到set里记录区间答案，然后像平衡树/线段树维护最大子串和那样，记录上边来的最长链，下边来的最长链，左边来的最长链，右边来的最长链……，以及当前结点保存的边长等信息。如果链是从原树的祖先方向来的，维护时候要加那个被迁移的边权，如果是从子孙方向来的，因为下面加过了所以不用再加……

之后花式更新即可。

那个边扔到set里的操作，有人说是传说中的虚边……ダメだ　 知らないだ……

n*(logn)^2卡常预警。据说是因为树浅时set多但LCT操作少，而树深时LCT操作多但set小，均摊下来不会超时。

　　↑不会算复杂度的我只能把这当成玄学

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
using namespace std;
const int INF=1e8;
const int mxn=200010;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int fir(multiset<int> &s){return s.size()? *s.rbegin():-INF;}
int sec(multiset<int> &s){return s.size()>1? *(++s.rbegin()):-INF;}
struct edge{int v,nxt,w;}e[mxn<<1];
int hd[mxn],mct=0;
void add_edge(int u,int v,int w){
    e[++mct].v=v;e[mct].nxt=hd[u];e[mct].w=w;hd[u]=mct;return;
}
int col[mxn],w[mxn];
struct LCT{
    multiset<int>chain[mxn],pt[mxn];
    int ch[mxn][2],fa[mxn];
    int len[mxn],L[mxn],R[mxn],mians[mxn],sum[mxn],ans;
    bool rev[mxn];
    void init(int n){for(int i=0;i<=n;i++)L[i]=R[i]=mians[i]=-INF;}
    inline bool isroot(int x){return ((ch[fa[x]][0]!=x) && (ch[fa[x]][1]!=x));}
    void pushup(int x){
        int lc=ch[x][0],rc=ch[x][1];
        sum[x]=sum[lc]+sum[rc]+len[x];
        int cmi=max(w[x],fir(chain[x]));
//        printf("tst %d:  lc:%d  rc:%d sum:%d  cmi:%d
",x,lc,rc,sum[x],cmi);
//        printf("     L rc:%d  R lc:%d
",L[rc],R[lc]);
        int La=max(cmi,R[lc]+len[x]);//左子树(深度更浅的地方) 
        int Ra=max(cmi,L[rc]);//右子树
        L[x]=max(L[lc],sum[lc]+len[x]+Ra);
        R[x]=max(R[rc],sum[rc]+La);
//        printf("     L:%d  R:%d
",L[x],R[x]);
        mians[x]=max(mians[lc],mians[rc]);
        mians[x]=max(mians[x],max(R[lc]+len[x]+Ra,L[rc]+La));
        mians[x]=max(mians[x],fir(pt[x]));
        mians[x]=max(mians[x],fir(chain[x])+sec(chain[x]));
        if(w[x]==0)mians[x]=max(w[x],max(mians[x],fir(chain[x])));
        return;
    }
    void rotate(int x){
        int y=fa[x],z=fa[y],lc,rc;
        if(ch[y][0]==x)lc=0;else lc=1;rc=lc^1;
        if(!isroot(y))ch[z][ch[z][1]==y]=x;
        fa[x]=z;fa[y]=x;
        ch[y][lc]=ch[x][rc];fa[ch[x][rc]]=y;
        ch[x][rc]=y;
        pushup(y);
        return;
    }
//    int st[mxn],top;
    void Splay(int x){
        while(!isroot(x)){
            int y=fa[x],z=fa[y];
            if(!isroot(y)){
                if((ch[z][0]==y)^(ch[y][0]==x))rotate(x);
                else rotate(y);
            }
            rotate(x);
        }
        pushup(x);
    }
    void access(int x){
        int y=0;
        for(;x;x=fa[x]){
            Splay(x);
            if(ch[x][1]){
                pt[x].insert(mians[ch[x][1]]);
                chain[x].insert(L[ch[x][1]]);
            }
            if(y){
                pt[x].erase(pt[x].find(mians[y]));
                chain[x].erase(chain[x].find(L[y]));
            }
            ch[x][1]=y;
            pushup(x);
            y=x;
        }
    }
    void change(int x){
        access(x);Splay(x);
        col[x]^=1;if(!col[x])w[x]=0;else w[x]=-INF;
        pushup(x);
        ans=mians[x];
    }
/*    void query(int x){
        access(x);Splay(x);
        pushup(x);
        ans=mians[x];
    }*/
}LT;
void DFS(int u,int fa){
    for(int i=hd[u];i;i=e[i].nxt){
        if(e[i].v==fa)continue;int v=e[i].v;
        LT.fa[v]=u;
        LT.len[v]=e[i].w;
        DFS(v,u);
        LT.chain[u].insert(LT.L[v]);
        LT.pt[u].insert(LT.mians[v]);
    }
    LT.pushup(u);
    return;
}
int n,Q;
int main(){
    int i,j,u,v,vl;
    n=read();
    for(i=1;i<n;i++){
        u=read();v=read();vl=read();
        add_edge(u,v,vl);
        add_edge(v,u,vl);
    }
    LT.init(n);
//    for(i=1;i<=n;i++)col[i]=0;
    DFS(1,0);
    LT.ans=LT.mians[1];
//    printf("LT:%d
",LT.ans);
    Q=read();char op[2];
    while(Q--){
        scanf("%s",op);
        if(op[0]=='C')v=read(),LT.change(v);
        else{
//            LT.query(v);
            if(LT.ans<0)printf("They have disappeared.
");
            else printf("%d
",LT.ans);
        }
    }
    return 0;
}