例題3.20 圖詢問 LA5031，3.20la5031

例題3.20 圖詢問 LA5031，3.20la5031

1.題目描述：點擊開啟連結

2.解題思路：本題利用Treap樹實現的名次樹來完成這三種操作。由於操作比較複雜，因此我們利用離線演算法來解決。可以實現把所有的D操作執行完，得到剩下的圖，接著按照逆序逐步插入邊，並在恰當的時機執行Q操作和C操作。用一棵名次樹維護一個連通分量的點權，則C操作對應於名次樹的一次修改操作（可以用一次刪除和一次插入來實現），Q操作對應Kth操作，而執行D操作時，如果兩個端點已知是同一個連通分量則無影響，否則將2個端點對應的2棵名次樹合并。這裡我們利用啟發學習法合并，讓結點數較小的樹合并到結點數較多的樹中，假設結點數分別為n1,n2，那麼時間複雜度為O(n1*logn2)。由於樹的結點總數不超過n，任意結點至多移動了log2N 次，每次移動需要logN的時間，因此總的時間複雜度為O(N(logN)^2)。

3.代碼：

#include<iostream>#include<algorithm>#include<cassert>#include<string>#include<sstream>#include<set>#include<bitset>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<cctype>#include<functional>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define me(s)  memset(s,0,sizeof(s))#define rep(i,n) for(int i=0;i<(n);i++)typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;//typedef pair <int, int> P;struct Node {  Node *ch[2]; // 左右子樹  int r; // 隨機優先順序  int v; // 值  int s; // 結點總數  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函數，因為可能會有相同結點    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();}const int maxc = 500000 + 10;struct Command {  char type;  int x, p; // 根據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];// 並查集相關int pa[maxn];int findset(int x) { return pa[x] != x ? pa[x] = findset(pa[x]) : x; }// 名次樹相關Node* root[maxn]; // Treapint 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;}// 主程式相關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));    // 讀命令    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 };    }    // 最終的圖    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\n", ++kase, query_tot / (double)query_cnt);  }  return 0;}

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