HDU 1512 Monkey King (左偏树+并查集)

题意：在一个森林里住着N(N<=10000)只猴子。在一开始，他们是互不认识的。但是随着时间的推移，猴子们少不了争斗，但那只会发生在互不认识

(认识具有传递性)的两只猴子之间。争斗时，两只猴子都会请出他认识的猴子里最强壮的一只(有可能是他自己)进行争斗。争斗后，这两只猴子就互相认识。

每个猴子有一个强壮值，但是被请出来的那两只猴子进行争斗后，他们的强壮值都会减半(例如10会减为5，5会减为2)。现给出每个猴子的初始强壮值，

给出M次争斗，如果争斗的两只猴子不认识，那么输出争斗后两只猴子的认识的猴子里最强壮的猴子的强壮值，否则输出 -1。

析：我们可以通过并查集来判断是不是同一群，然后由于要改值和合并，要有一个合适的数据结构来维护，很容易想到是用左偏树，不过好像Splay也行。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100000 + 10;
const int mod = 1e6 + 10;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int p[maxn];
int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); }
struct Node{
  int key, l, r, fa, d;
};
Node tr[maxn];

int iroot(int i){
  if(i == -1)  return i;
  while(tr[i].fa != -1)  i = tr[i].fa;
  return i;
}

int Merge(int rx, int ry){
  if(rx == -1)  return ry;
  if(ry == -1)  return rx;
  if(tr[rx].key < tr[ry].key)  swap(rx, ry);
  int r = Merge(tr[rx].r, ry);
  tr[rx].r = r;  tr[r].fa = rx;
  if(tr[r].d > tr[tr[rx].l].d)  swap(tr[rx].r, tr[rx].l);
  if(tr[rx].r == -1)  tr[rx].d = 0;
  else tr[rx].d = tr[tr[rx].r].d + 1;
  return rx;
}

int Insert(int i, int key, int root){
  tr[i].key = key;
  tr[i].l = tr[i].r = tr[i].fa = -1;
  tr[i].d = 0;
  return Merge(root, i);
}

int del(int i){
  if(i == -1)  return -1;
  int l = tr[i].l, r = tr[i].r, y = tr[i].fa;
  tr[i].l = tr[i].r = tr[i].fa = -1;
  int x;
  tr[x = Merge(l, r)].fa = y;
  if(y != -1 && tr[y].l == i)  tr[y].l = x;
  if(y != -1 && tr[y].r == i)  tr[y].r = x;
  for( ; y != -1; x = y, y = tr[y].fa){
    if(tr[tr[y].l].d < tr[tr[y].r].d)  swap(tr[y].l, tr[y].r);
    if(tr[y].d == tr[tr[y].r].d + 1)  break;
    tr[y].d = tr[tr[y].r].d + 1;
  }
  if(x != -1)  return iroot(x);
  return iroot(y);
}

Node top(int root){ return tr[root]; }


int add(int i){
  if(i == -1)  return i;
  if(tr[i].l == -1 && tr[i].r == -1 && tr[i].fa == -1){
    tr[i].key /= 2;
    return i;
  }
  int key = tr[i].key / 2;
  int rt = del(i);
  return Insert(i, key, rt);
}

void init(){
  for(int i = 1; i <= n; ++i){
    p[i] = i;
    scanf("%d", &tr[i].key);
    tr[i].l = tr[i].r = tr[i].fa = -1;
    tr[i].d = 0;
  }
}

int main(){
  while(scanf("%d", &n) == 1){
    init();
    scanf("%d", &m);
    while(m--){
      int u, v;
      scanf("%d %d", &u, &v);
      int x = Find(u);
      int y = Find(v);
      if(x == y){ printf("-1
");  continue; }
      p[y] = x;
      int rt = iroot(u);
      int rt1 = iroot(v);
      rt = add(rt);   rt1 = add(rt1);
      int t = Merge(rt, rt1);
      printf("%d
", tr[t].key);
    }
  }
  return 0;
}

HDU 1512 Monkey King (左偏树+并查集)

相关推荐