Many people have covered this topic in great detail, so I won't repeat it.
To sum up, this can be regarded as a "side-weighted" union, for which we can record the relationship on the edge of the array separately, and then maintain the relationship while finding and unite.
So in this question, we use a d array to record the number of battleships from the current battleship to the battleship in the front of this column of battleships,
Use a size array (in the program to avoid the keyword replaced by siz) to record how many battleships are followed by the current battleship (update d for the convenience of M operation).
Refer to the code for specific maintenance methods.
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int MAXT = 5e5 + 20;
4 const int MAXN = 3e4 + 20;
5
6 namespace ufs
7 {
8 int fa[MAXN], d[MAXN], siz[MAXN];
9
10 void init()
11 {
12 for(int i = 1; i <= MAXN; i++)
13 fa[i] = i, siz[i] = 1;
14 }
15
16 int get(int x)
17 {
18 if(x == fa[x]) return x;
19 int root = get(fa[x]);
20 d[x] += d[fa[x]];
21 return fa[x] = root;
22 }
23
24 inline void unite(int x, int y)
25 {
26 x = get(x), y = get(y);
27 fa[x] = y, d[x] = siz[y];
28 siz[y] += siz[x];
29 }
30 }
31
32
33 int main()
34 {
35 //freopen("p1196.txt", "r", stdin);
36 int T;
37 cin>>T;
38 using namespace ufs;
39 init();
40
41 char opt;int i, j;
42 while(T--)
43 {
44 scanf(" %c ", &opt);
45 if(opt == 'M')
46 {
47 scanf("%d%d", &i, &j);
48 unite(i, j);
49 }
50 else
51 {
52 scanf("%d%d", &i, &j);
53 if(get(i) != get(j)) puts("-1");
54 else printf("%d\n", abs(d[i] - d[j]) - 1);
55 }
56 }
57 return 0;
58 }
Note a detail: don't forget that d[ x ] holds the number of warships before x, and d[ y ] holds the number of warships before y, so the number of warships between x and y should be the absolute value of the difference between the two Subtract 1 again.