【treap tree】 HDOJ 3726 && LA 5031 && UVA 1479 Graph and Queries

在线转离线。。先构建出最终的图，然后逆着处理。。不过有几个要注意的地方。。。答案超int，要用long long。。逆着处理的时候C操作要反着操作。。求第k大的时候，k可能是0或负数，也可能超过集合元素个数。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 20005
#define eps 1e-10
#define mod 1000000009
#define INF 99999999
#define lowbit(x) (x&(-x))
//#define lson o<<1, L, mid
//#define rson o<<1 | 1, mid+1, R
typedef long long LL;
//typedef int LL;
using namespace std;

int value[maxn], n, m;
char s[100];
struct Edge
{
int u, v, done;
}edges[60005];
struct opr
{
int kk, x, y;
}op[1000005];
int opp, cnt;
LL ans;
int f[maxn], num[maxn];
struct node
{
int key, v, s;
struct node *ch[2];
int cmp(int x) const {
if(x == v) return -1;
return x < v ? 0 : 1;
}
void maintain(void)
{
s = ch[0]->s + ch[1]->s + 1;
}
}*null, *root[maxn];

void read(void)
{
opp = 0;
for(int i = 1; i <= n; i++) scanf("%d", &value[i]);
for(int i = 1; i <= m; i++) scanf("%d%d", &edges[i].u, &edges[i].v), edges[i].done = 0;
while(scanf("%s", s)!=EOF) {
if(s[0] == 'E') break;
if(s[0] == 'D') op[opp].kk = 0, scanf("%d", &op[opp++].x), edges[op[opp-1].x].done = 1;
if(s[0] == 'Q') op[opp].kk = 1, scanf("%d%d", &op[opp].x, &op[opp].y), opp++;
if(s[0] == 'C') op[opp].kk = 2, scanf("%d%d", &op[opp].x, &op[opp].y), swap(op[opp].y, value[op[opp].x]), opp++;
}
}
void init(void)
{
int i;
ans = cnt = 0;
for(i = 0; i < maxn; i++) num[i] = 1;
for(i = 0; i < maxn; i++) f[i] = i;
null = new node;
null->s = null->v = 0;
for(i = 0; i <= n; i++) {
root[i] = new node;
root[i] = null;
}
}
int find(int x)
{
return f[x] == x ? f[x] : find(f[x]);
}
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;
o->key = rand(), o->v = x;
o->s = 1, o->ch[0] = o->ch[1] = null;
return;
}
int d = x < o->v ? 0 : 1;
insert(o->ch[d], x);
if(o->key < o->ch[d]->key) rotate(o, d^1);
o->maintain();
}
void remove(node* &o, int x)
{
int d = o->cmp(x);
if(d == -1) {
if(o->ch[0] != null && o->ch[1] != null) {
int d2 = (o->ch[0]->key > o->ch[1]->key ? 1 : 0);
rotate(o, d2), remove(o->ch[d2], x);
}
else {
node *u = o;
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 || o->s < k || k <= 0) return 0;
if(o->ch[1]->s >= k) return kth(o->ch[1], k);
if(o->ch[1]->s + 1 == k) return o->v;
return kth(o->ch[0], k - o->ch[1]->s - 1);
}
void debug(node* o)
{
printf("FSDF  %d\n", o->v);
if(o->ch[0] != null) debug(o->ch[0]);
if(o->ch[1] != null) debug(o->ch[1]);
}
void removetree(node* &o, node* &p)
{
if(o == null) return;
if(o->ch[0] != null) removetree(o->ch[0], p);
if(o->ch[1] != null) removetree(o->ch[1], p);
insert(p, o->v);
delete o;
}
void updata(int a, int v)
{
int aa = find(a);
remove(root[aa], value[a]);
value[a] = v;
insert(root[aa], value[a]);
}
void query(int a, int k)
{
int aa = find(a);
ans += kth(root[aa], k);
}
void merge(int a, int b)
{
int aa = find(a), bb = find(b);
if(aa == bb) return;
if(num[aa] > num[bb]) {
removetree(root[bb], root[aa]);
f[bb] = aa, num[aa] += num[bb], num[bb] = 0;
root[bb] = null;
}
else {
removetree(root[aa], root[bb]);
f[aa] = bb, num[bb] += num[aa], num[aa] = 0;
root[aa] = null;
}
}
void work(void)
{
int i, aa, bb;
for(i = 1; i <= m; i++)
if(!edges[i].done) {
aa = find(edges[i].u);
bb = find(edges[i].v);
if(aa != bb) {
if(aa > bb) f[aa] = bb, num[bb] += num[aa], num[aa] = 0;
else f[bb] = aa, num[aa] += num[bb], num[bb] = 0;
}
}
for(i = 1; i <= n; i++) insert(root[find(i)], value[i]);
for(i = opp-1; i >= 0; i--) {
if(op[i].kk == 0) merge(edges[op[i].x].u, edges[op[i].x].v);
if(op[i].kk == 1) query(op[i].x, op[i].y), cnt++;
if(op[i].kk == 2) updata(op[i].x, op[i].y);
}
}
int main(void)
{
int _ = 0;
while(scanf("%d%d", &n, &m), n!=0 || m!=0) {
init();
read();
work();
printf("Case %d: %.6f\n", ++_, ans/(double)cnt);
}
return 0;
}