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[], w[], d[], l[], r[];

 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]] + ;
     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] = ;
     return merge(lc, rc);
 }

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