[Monkey King]

题目描述

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。

输入

There are several test cases, and each case consists of two parts.

First part: The first line contains an integer N(N<=100,000), which indicates the number of monkeys. And then N lines follows. There is one number on each line, indicating the strongness value of ith monkey(<=32768).

Second part: The first line contains an integer M(M<=100,000), which indicates there are M conflicts happened. And then M lines follows, each line of which contains two integers x and y, indicating that there is a conflict between the Xth monkey and Yth.

输出

For each of the conflict, output -1 if the two monkeys know each other, otherwise output the strongness value of the strongest monkey in all friends of them after the duel.

样例输入

5
20
16
10
10
4
5
2 3
3 4
3 5
4 5
1 5

样例输出

8
5
5
-1
10

 

删除就是把左子树合并到右子树，然后再把根节点除以2加入右子树中

 

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<cstdlib>
 #include<ctime>
 using namespace std;
 const int N=;
 int n;int fa[N];
 struct node{
     int x,dis;
     node *l,*r;
     int ldis(){return l?l->dis:;}
     int rdis(){return r?r->dis:;}
 }a[N];
 int gi()
 {
     int str=;char ch=getchar();
     while(ch>'' || ch<'')ch=getchar();
     while(ch>='' && ch<='')str=str*+ch-'',ch=getchar();
     return str;
 }
 int find(int x){
     return x==fa[x]?x:fa[x]=find(fa[x]);
 }
 node *root[N],*pos=a;
 void newnode(node *&p,int key)
 {
     p=pos++;
     p->l=p->r=NULL;
     p->dis=;p->x=key;
 }
 node *merge(node *p,node *q)
 {
     if(!p || !q)return p?p:q;
     if(p->x<q->x)swap(p,q);
     p->r=merge(p->r,q);
     if(p->ldis()<p->rdis())swap(p->l,p->r);
     p->dis=p->rdis()+;
     return p;
 }
 void Delet(int t)
 {
     node *R=root[t]->r;
     node *L=root[t]->l;
     node *rt=root[t];
     rt->dis=;rt->l=rt->r=NULL;rt->x>>=;
     root[t]=merge(R,L);
     root[t]=merge(root[t],rt);
 }
 void work()
 {
     int x;
     for(int i=;i<=n;i++)
     {
         fa[i]=i;x=gi();
         newnode(root[i],x);
     }
     int m=gi(),y;
     while(m--)
     {
         x=gi();y=gi();
         if(find(x)==find(y)){
             printf("-1\n");
             continue;
         }
         else{
             int xx=find(x),yy=find(y);
             Delet(xx);Delet(yy);
             fa[yy]=xx;
             root[xx]=merge(root[xx],root[yy]);
             printf("%d\n",root[xx]->x);
         }
     }
     return ;
 }
 int main()
 {
     while(~scanf("%d",&n)){
         work();
         pos=a;
     }
     return ;
 }