Monkey King

Posted 2020-09-11 蒟蒻zht的博客

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。

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 
 6 int n, m;
 7 int f[100010], w[100010], d[100010], l[100010], r[100010];
 8 
 9 inline int merge(int x, int y)
10 {
11     if(!x || !y) return x + y;
12     if(w[x] < w[y]) swap(x, y);
13     r[x] = merge(r[x], y);
14     f[r[x]] = x;
15     if(d[l[x]] < d[r[x]]) swap(l[x], r[x]);
16     d[x] = d[r[x]] + 1;
17     return x;
18 }
19 
20 inline int find(int x)
21 {
22     return x == f[x] ? x : f[x] = find(f[x]);
23 }
24 
25 inline int pop(int x)
26 {
27     int lc = l[x], rc = r[x];
28     f[lc] = lc;
29     f[rc] = rc;
30     l[x] = r[x] = d[x] = 0;
31     return merge(lc, rc);
32 }
33 
34 int main()
35 {
36     int i, j, x, y, fx, fy, x1, y1;
37     while(~scanf("%d", &n))
38     {
39         for(i = 1; i <= n; i++)
40         {
41             scanf("%d", &w[i]);
42             f[i] = i;
43             l[i] = 0;
44             r[i] = 0;
45             d[i] = 0;
46         }
47         scanf("%d", &m);
48         for(i = 1; i <= m; i++)
49         {
50             scanf("%d %d", &x, &y);
51             fx = find(x);
52             fy = find(y);
53             if(fx == fy)
54             {
55                 printf("-1\n");
56                 continue;
57             }
58             x1 = pop(fx);
59             w[fx] /= 2;
60             x1 = merge(x1, fx);
61             y1 = pop(fy);
62             w[fy] /= 2;
63             y1 = merge(y1, fy);
64             printf("%d\n", w[merge(x1, y1)]);
65         }
66     }
67     return 0;
68 }

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