SPOJ-QTREE4 Query on a tree IV(边分治)

QTREE4 - Query on a tree IV

#tree

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:31 2 11 3 17AC 1AC 2AC 3AOutput:220They have disappeared.

题意：

给定一棵树，节点有黑白两种颜色，有正负的边权。有两种操作：

①修改反转某个节点的颜色；

②询问树上最远的两个白色节点的距离。

#include<bits/stdc++.h>using namespace std;const int MX = 2e5 + 5;const int MXE = 4e6 + 5;struct Edge {    int v, w, nxt, pre;} E[MXE], edge[MXE];int Head[MX], head[MX], rear, tot, tail[MX];int mark[MX], sz[MX];int N, n, cnt, rt, midedge, Max;void init() {    memset(head, -1, sizeof(head));    tot = 0;}void INIT() {    memset(Head, -1, sizeof(Head));    rear = 0;}void add(int u, int v, int w) {    edge[tot].v = v;    edge[tot].w = w;    edge[tot].nxt = head[u];    head[u] = tot++;}void ADD(int u, int v, int w) {    E[rear].v = v;    E[rear].w = w;    E[rear].nxt = Head[u];    Head[u] = rear++;}void Delete(int u, int i) {    if (Head[u] == i) Head[u] = E[i].nxt;    else E[E[i].pre].nxt = E[i].nxt;    if (tail[u] == i) tail[u] = E[i].pre;    else E[E[i].nxt].pre = E[i].pre;}//保证每个点的度不超过3void build(int u, int fa) {    int father = 0;    for (int i = head[u]; ~i; i = edge[i].nxt) {        int v = edge[i].v, w = edge[i].w;        if (v == fa) continue;        if (father == 0) { //还没有增加子节点，直接连上            ADD(u, v, w); ADD(v, u, w);            father = u;            build(v, u);        } else {        //已经有一个子节点，则创建一个新节点，把v连在新节点上            mark[++N] = 0;            ADD(N, father, 0); ADD(father, N, 0);            father = N;            ADD(v, father, w); ADD(father, v, w);            build(v, u);        }    }}//nxt是下一条边的编号，pre是上一条边的编号void get_pre() {    memset(tail, -1, sizeof(tail));    for (int i = 1; i <= N; i++) {        for (int j = Head[i]; ~j; j = E[j].nxt) {            E[j].pre = tail[i];            tail[i] = j;        }    }}//重建一个图void rebuild() {    INIT();    N = n;    for (int i = 1; i <= N; i++) mark[i] = 1;    build(1, 0);    get_pre();    init();}struct point {    int u, dis;    point() {}    point(int _u, int _dis) {        u = _u; dis = _dis;    }    bool operator<(const point& _A)const {        return dis < _A.dis;    }};struct node {    int rt, midlen, ans; //根节点，中心边，答案(最长树链)    int ls, rs;         //左右子树编号    priority_queue<point>q;} T[2*MX];//搜索每个子树大小void dfs_size(int u, int fa, int dir) {    add(u, rt, dir);    //如果是白点，则压入根节点rt的队列，dist为到根的距离    //队列中的点用来pt的父亲树的更新，pt节点不需要用到，    //因此T[pt].rt是父亲树的中心边上的点    if (mark[u]) T[rt].q.push(point(u, dir));    sz[u] = 1;    for (int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v, w = E[i].w;        if (v == fa) continue;        dfs_size(v, u, dir + w);        sz[u] += sz[v];    }}//找中心边void dfs_midedge(int u, int code) {    if (max(sz[u], sz[T[rt].rt] - sz[u]) < Max) {        Max = max(sz[u], sz[T[rt].rt] - sz[u]);        midedge = code;    }    for (int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if (i != (code ^ 1)) dfs_midedge(v, i);    }}//更新void PushUP(int rt) {    T[rt].ans = -1;    while (!T[rt].q.empty() && mark[T[rt].q.top().u] == 0) T[rt].q.pop();//弹出黑点    int ls = T[rt].ls, rs = T[rt].rs; //ls为左儿子，rs为右儿子    if (ls == 0 && rs == 0) { //没有左右儿子        if (mark[T[rt].rt])T[rt].ans = 0;    } else {        if (T[ls].ans > T[rt].ans) T[rt].ans = T[ls].ans; //如果左儿子的结果大于右儿子        if (T[rs].ans > T[rt].ans) T[rt].ans = T[rs].ans; //如果右儿子的结果大于左儿子        if (!T[ls].q.empty() && !T[rs].q.empty()) //穿过中心边的            T[rt].ans = max(T[rt].ans, T[ls].q.top().dis + T[rs].q.top().dis + T[rt].midlen);    }}void DFS(int id, int u) {    rt = id; Max = N; midedge = -1;    T[id].rt = u;    dfs_size(u, 0, 0);    dfs_midedge(u, -1);    if (~midedge) {        //中心边的左右2点        int p1 = E[midedge].v;        int p2 = E[midedge ^ 1].v;        //中心边长度        T[id].midlen = E[midedge].w;        //左右子树        T[id].ls = ++cnt;        T[id].rs = ++cnt;        //删除中心边        Delete(p1, midedge ^ 1);        Delete(p2, midedge);        DFS(T[id].ls, p1);        DFS(T[id].rs, p2);    }    PushUP(id);}void update(int u) {    mark[u] ^= 1;    for (int i = head[u]; ~i; i = edge[i].nxt) {        int v = edge[i].v, w = edge[i].w;        if (mark[u] == 1) T[v].q.push(point(u, w));        PushUP(v);    }}int main() {    //freopen("in.txt","r",stdin);    scanf("%d", &n);    init();    for (int i = 1, u, v, w; i < n; i++) {        scanf("%d%d%d", &u, &v, &w);        add(u, v, w); add(v, u, w);    }    rebuild();    DFS(cnt = 1, 1);    char op[2]; int m, x;    scanf("%d", &m);    while (m--) {        scanf("%s", op);        if (op[0] == 'A') {            if (T[1].ans == -1) printf("They have disappeared.\n");            else printf("%d\n", T[1].ans);        } else {            scanf("%d", &x);            update(x);        }    }    return 0;}