Monkey King（左偏树）

洛谷传送门

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>

using namespace std;

int n, m;
int f[100010], w[100010], d[100010], l[100010], r[100010];

inline int merge(int x, int y)
{
    if(!x || !y) return x + y;
    if(w[x] < w[y]) swap(x, y);
    r[x] = merge(r[x], y);
    f[r[x]] = x;
    if(d[l[x]] < d[r[x]]) swap(l[x], r[x]);
    d[x] = d[r[x]] + 1;
    return x;
}

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

inline int pop(int x)
{
    int lc = l[x], rc = r[x];
    f[lc] = lc;
    f[rc] = rc;
    l[x] = r[x] = d[x] = 0;
    return merge(lc, rc);
}

int main()
{
    int i, j, x, y, fx, fy, x1, y1;
    while(~scanf("%d", &n))
    {
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &w[i]);
            f[i] = i;
            l[i] = 0;
            r[i] = 0;
            d[i] = 0;
        }
        scanf("%d", &m);
        for(i = 1; i <= m; i++)
        {
            scanf("%d %d", &x, &y);
            fx = find(x);
            fy = find(y);
            if(fx == fy)
            {
                printf("-1
");
                continue;
            }
            x1 = pop(fx);
            w[fx] /= 2;
            x1 = merge(x1, fx);
            y1 = pop(fy);
            w[fy] /= 2;
            y1 = merge(y1, fy);
            printf("%d
", w[merge(x1, y1)]);
        }
    }
    return 0;
}