HDU 1512 MonkeyKing

//通过这题知道了可并堆这个东西.其实挺好理解的，体会到了
//数据结构的强大啊！！！

#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXN = 100100;
int father[MAXN];
struct Monkey
{
    int l,r;
    int dis;
    int strong;
}LTree[MAXN];
int find(int x){
    if(x != father[x])
        father[x] = find(father[x]);
    return father[x];
}
int merge(int x,int y){                        //返回合并后的根
    if( x == 0 )
        return y;
    if( y == 0)
        return x;
    if( LTree[x].strong < LTree[y].strong)            //大顶堆
        swap(x,y);
    LTree[x].r = merge(LTree[x].r,y);            //递归合并右子树和Y
    int l = LTree[x].l , r = LTree[x].r;
    father[r] = x;                        //更新T右子树的根 
    if(LTree[l].dis < LTree[r].dis)                //维护堆性质
        swap(LTree[x].l,LTree[x].r);
    if(LTree[x].r == 0)                    //如果没有右子树 则距离为0
        LTree[x].dis = 0;
    else
        LTree[x].dis = LTree[LTree[x].r].dis + 1;
    return x;
}
int del(int x){            //返回删除根以后左右子树的合并的根
    int l,r;
    l = LTree[x].l;
    r = LTree[x].r;
    father[l] = l;
    father[r] = r;
    LTree[x].l = LTree[x].r = LTree[x].dis = 0;
    return merge(l,r);
}
void solve(int x,int y){
    LTree[x].strong /= 2;
    LTree[y].strong /= 2;
    //问每次PK以后，当前这个群体里力量最大的猴子的力量是多少。
    int left,right;
    left = del(x);
    right = del(y);
    left = merge(left,x);
    right = merge(right,y);
    left = merge(left,right);
    printf("%d\n",LTree[left].strong);
}
int main(){
    //freopen("1005.txt","r",stdin);
    int n,m,i,x,y;
    while(scanf("%d",&n) == 1){
        for(i = 1; i <= n; i++){
            scanf("%d",&LTree[i].strong);
            LTree[i].l = 0;
            LTree[i].r = 0;
            LTree[i].dis = 0;
            father[i] = i;            //起始已自己为父亲
        }
        scanf("%d",&m);
        for(i = 1; i <= m; i++){
            scanf("%d%d",&x,&y);
            int fx = find(x),fy = find(y);
            if(fx == fy){
                printf("-1\n");
            } else {
                solve(fx,fy);
            }
        }
    }
    return 0;
}