hdu1512 Monkey King（并查集，左偏堆）

/*
每次给出要争吵的猴子a和b，用并查集判断如果他们是朋友输出-1
如果不是，找出a，b在的堆的根A，B，分别合并A，B的左右孩子，再合并一下。
之后把A，B的数据更改一下：权值除以2，左右孩子设为0，再插入到堆中即可。
最后输出堆顶。
*/ 
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100005;
int m,n;
int set[N];
struct node
{
    int l,r,dis,key;
}tree[N];

int find(int x) {return set[x] == x ? x : set[x] = find(set[x]);}

int merge(int a,int b)
{
    if(!a)
        return b;
    if(!b)
        return a;
    if(tree[a].key < tree[b].key)//大堆
        swap(a,b);
    tree[a].r = merge(tree[a].r,b);
    set[tree[a].r] = a;//并查
    if(tree[tree[a].l].dis < tree[tree[a].r].dis)
        swap(tree[a].l,tree[a].r);
    if(tree[a].r)
        tree[a].dis = tree[tree[a].r].dis + 1;
    else
        tree[a].dis = 0;
    return a;
}

int pop(int a)
{
    int l = tree[a].l;
    int r = tree[a].r;
    set[l] = l;//因为要暂时删掉根，所以左右子树先作为根
    set[r] = r;
    tree[a].l = tree[a].r = tree[a].dis = 0;
    return merge(l,r);
}

int nextint()
{
    char c;
    int ret = 0;
    while((c = getchar()) > ‘9‘ || c < ‘0‘)
        ;
    ret = c - ‘0‘;
    while((c = getchar()) >= ‘0‘ && c <= ‘9‘)
        ret = ret * 10 + c - ‘0‘;
    return ret;
}

void print(int a)
{
    if(!a)
        return;
    print(a/10);
    putchar(a%10 + ‘0‘);
}

int main()
{
    int a,b,i;
    while(~scanf("%d",&n))
    {
        for(i = 1;i <= n;i ++)
        {;
            tree[i].key = nextint();
            set[i] = i;
            tree[i].l = tree[i].r = tree[i].dis = 0;
        }
        m = nextint();
        while(m --)
        {
            a = nextint();
            b = nextint();
            int ra = find(a);
            int rb = find(b);
            if(ra == rb)
                printf("-1\n");
            else
            {
                int rra = pop(ra);//ra左右子树合并
                tree[ra].key /= 2;
                ra = merge(rra,ra);//重新插入ra 找到合适的位置
                int rrb = pop(rb);
                tree[rb].key /= 2;
                rb = merge(rrb,rb);
                print(tree[merge(ra,rb)].key);
                putchar(10);
            }
        }
    }
    return 0;
}