【题目链接】http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=20332
【PS】算法入门经典Treap的例题，树上讲得非常仔细，在P234。再大致说一下这一个题，很经典。
【题意】给定无向图，有三种操作，①删除第i条边②查询节点x所在的连通分量节点中第K大的权值③改变节点x的权值。现在问你最终的查询结果的平均值
【解题方法】先读入所有的操作，删除所有边得到最终的图，然后反过来操作，在适合的时候加上边，在适合的时候改变点的权值，然后用名次树来维护每个连通分量，加边的时候就用并查集判一下就可以了，然后再将两棵树合并，这里用了启发式合并。
【复杂度】O(n*logn*logn)
【代码君】
//
//Created by just_sort 2016/10/14
//Copyright (c) 2016 just_sort.All Rights Reserved
//
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 22222;
const int maxm = 66666;
const int maxq = 555555;
struct node{
node *ch[2];
int r,v,s;
node(int v) : v(v) { ch[0] = ch[1] = NULL; r = rand(); s = 1;}
bool operator < (const node &rhs) const{
return r < rhs.r;
}
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);
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);
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();
}
int Kth(node *o, int k)
{
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);
}
struct Q{
char type;
int x, p;
}q[maxq];
int n,m;
int from[maxm],to[maxm],removed[maxm],weight[maxm];
int pa[maxn];
int Find(int x)
{
if(x == pa[x]) return x;
return pa[x] = Find(pa[x]);
}
node *root[maxn]; //Treap
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;
}
void addedge(int x)
{
int u = Find(from[x]), v = Find(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 cnt;
LL tot;
void query(int x,int k){
cnt++;
tot += Kth(root[Find(x)], k);
}
void change_weight(int x,int v)
{
int u = Find(x);
remove(root[u], weight[x]);
insert(root[u], v);
weight[x] = v;
}
void work()
{
int ks = 1;
while(scanf("%d%d", &n,&m)!=EOF)
{
if(n == 0 && m == 0) break;
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));
int c = 0;
while(1)
{
char op[10];
int x, p = 0, v = 0;
scanf("%s", op);
if(op[0] == 'E') break;
scanf("%d", &x);
if(op[0] == 'D') removed[x] = 1;
if(op[0] == 'Q') scanf("%d", &p);
if(op[0] == 'C') {
scanf("%d", &v);
p = weight[x];
weight[x] = v;
}
q[c++] = (Q){op[0], x, p};
}
//删除之后最终的图
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]) addedge(i);
//反向操作
tot = cnt = 0;
for(int i = c - 1; i >= 0; i--){
if(q[i].type == 'D') addedge(q[i].x);
if(q[i].type == 'Q') query(q[i].x, q[i].p);
if(q[i].type == 'C') change_weight(q[i].x, q[i].p);
}
printf("Case %d: %.6f\n",ks++,tot/(double)cnt);
}
}
int main()
{
work();
return 0;
}