uvalive 5031 Graph and Queries 名次树+Treap

题意：给你个点m条边的无向图，每个节点都有一个整数权值。你的任务是执行一系列操作。操作分为3种。。。

思路：本题一点要逆向来做，正向每次如果删边，复杂度太高。逆向到一定顺序的时候添加一条边更容易。详见算法指南P235。

#include<cstdlib>

struct Node
{
    Node *ch[2]; // 左右子树
    int r; // 随机优先级
    int v; // 值
    int s; // 结点总数
    Node(int v):v(v)
    {
        ch[0] = ch[1] = NULL;
        r = rand();
        s = 1;
    }
    int cmp(int x) const
    {
        if (x == v) return -1;
        return x < v ? 0 : 1;
    }
    void maintain()
    {
        s = 1;
        if(ch[0] != NULL) s += ch[0]->s;
        if(ch[1] != NULL) s += ch[1]->s;
    }
};

void rotate(Node* &o, int d)
{
    Node* k = o->ch[d^1];
    o->ch[d^1] = k->ch[d];
    k->ch[d] = o;
    o->maintain();
    k->maintain();
    o = k;
}

void insert(Node* &o, int x)
{
    if(o == NULL) o = new Node(x);
    else
    {
        int d = (x < o->v ? 0 : 1); // 不要用cmp函数，因为可能会有相同结点
        insert(o->ch[d], x);
        if(o->ch[d]->r > o->r) rotate(o, d^1);
    }
    o->maintain();
}

void remove(Node* &o, int x)
{
    int d = o->cmp(x);
    int ret = 0;
    if(d == -1)
    {
        Node* u = o;
        if(o->ch[0] != NULL && o->ch[1] != NULL)
        {
            int d2 = (o->ch[0]->r > o->ch[1]->r ? 1 : 0);
            rotate(o, d2);
            remove(o->ch[d2], x);
        }
        else
        {
            if(o->ch[0] == NULL) o = o->ch[1];
            else o = o->ch[0];
            delete u;
        }
    }
    else
        remove(o->ch[d], x);
    if(o != NULL) o->maintain();
}

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

const int maxc = 500000 + 10;
struct Command
{
    char type;
    int x, p; // 根据type, p代表k或者v
} commands[maxc];

const int maxn = 20000 + 10;
const int maxm = 60000 + 10;
int n, m, weight[maxn], from[maxm], to[maxm], removed[maxm];

// 并查集相关
int pa[maxn];
int findset(int x)
{
    return pa[x] != x ? pa[x] = findset(pa[x]) : x;
}

// 名次树相关
Node* root[maxn]; // Treap

int kth(Node* o, int k)   // 第k大的值
{
    if(o == NULL || k <= 0 || k > o->s) return 0;
    int s = (o->ch[1] == NULL ? 0 : o->ch[1]->s);
    if(k == s+1) return o->v;
    else if(k <= s) return kth(o->ch[1], k);
    else return kth(o->ch[0], k-s-1);
}

void mergeto(Node* &src, Node* &dest)
{
    if(src->ch[0] != NULL) mergeto(src->ch[0], dest);
    if(src->ch[1] != NULL) mergeto(src->ch[1], dest);
    insert(dest, src->v);
    delete src;
    src = NULL;
}

void removetree(Node* &x)
{
    if(x->ch[0] != NULL) removetree(x->ch[0]);
    if(x->ch[1] != NULL) removetree(x->ch[1]);
    delete x;
    x = NULL;
}

// 主程序相关
void add_edge(int x)
{
    int u = findset(from[x]), v = findset(to[x]);
    if(u != v)
    {
        if(root[u]->s < root[v]->s)
        {
            pa[u] = v;
            mergeto(root[u], root[v]);
        }
        else
        {
            pa[v] = u;
            mergeto(root[v], root[u]);
        }
    }
}

int query_cnt;
long long query_tot;
void query(int x, int k)
{
    query_cnt++;
    query_tot += kth(root[findset(x)], k);
}

void change_weight(int x, int v)
{
    int u = findset(x);
    remove(root[u], weight[x]);
    insert(root[u], v);
    weight[x] = v;
}

int main()
{
    int kase = 0;
    while(scanf("%d%d", &n, &m) == 2 && n)
    {
        for(int i = 1; i <= n; i++) scanf("%d", &weight[i]);
        for(int i = 1; i <= m; i++) scanf("%d%d", &from[i], &to[i]);
        memset(removed, 0, sizeof(removed));

        // 读命令
        int c = 0;
        for(;;)
        {
            char type;
            int x, p = 0, v = 0;
            scanf(" %c", &type);
            if(type == 'E') break;
            scanf("%d", &x);
            if(type == 'D') removed[x] = 1;
            if(type == 'Q') scanf("%d", &p);
            if(type == 'C')
            {
                scanf("%d", &v);
                p = weight[x];
                weight[x] = v;
            }
            commands[c++] = (Command)
            {
                type, x, p
            };
        }

        // 最终的图
        for(int i = 1; i <= n; i++)
        {
            pa[i] = i;
            if(root[i] != NULL) removetree(root[i]);
            root[i] = new Node(weight[i]);
        }
        for(int i = 1; i <= m; i++) if(!removed[i]) add_edge(i);

        // 反向操作
        query_tot = query_cnt = 0;
        for(int i = c-1; i >= 0; i--)
        {
            if(commands[i].type == 'D') add_edge(commands[i].x);
            if(commands[i].type == 'Q') query(commands[i].x, commands[i].p);
            if(commands[i].type == 'C') change_weight(commands[i].x, commands[i].p);
        }
        printf("Case %d: %.6lf
", ++kase, query_tot / (double)query_cnt);
    }
    return 0;
}