例题3.20 图询问 LA5031

标签：treap   平衡树   名次树   

1.题目描述：点击打开链接

2.解题思路：本题利用Treap树实现的名次树来完成这三种操作。由于操作比较复杂，因此我们利用离线算法来解决。可以实现把所有的D操作执行完，得到剩下的图，接着按照逆序逐步插入边，并在恰当的时机执行Q操作和C操作。用一棵名次树维护一个连通分量的点权，则C操作对应于名次树的一次修改操作（可以用一次删除和一次插入来实现），Q操作对应Kth操作，而执行D操作时，如果两个端点已知是同一个连通分量则无影响，否则将2个端点对应的2棵名次树合并。这里我们利用启发式合并，让结点数较小的树合并到结点数较多的树中，假设结点数分别为n1,n2，那么时间复杂度为O(n1*logn2)。由于树的结点总数不超过n，任意结点至多移动了log2N 次，每次移动需要logN的时间，因此总的时间复杂度为O(N(logN)^2)。

3.代码：

#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();
}


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\n", ++kase, query_tot / (double)query_cnt);
  }
  return 0;
}

例题3.20 图询问 LA5031

标签：treap   平衡树   名次树   

原文地址：http://blog.csdn.net/u014800748/article/details/48119623