银河英雄传说 洛谷1196 并查集

　　有点有趣的并查集问题，类似于食物链而略简单
　　使用sum[i]表示以i为祖先的并查集的元素个数（不包含i）， 以dis[i]表示i到其祖先的距离，每次getfather时根据x的直接父亲来更新x的dis值，每次修改操作时用sum[y]更新dis[x]，用sum[x]更新sum[y]，最后输出时直接利用两个点的dis值得解

1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4
5
6 const int maxn = 30000 + 500;
7 int father[maxn], dis[maxn];
8 int sum[maxn];
9 char c;
10 int x, y;
11 int t;
12
13 int getfather(int x) {
14     if (father[x] == x) return (x);
15     int cur = father[x];
16     father[x] = getfather(father[x]);
17     dis[x] = dis[x] + dis[cur];
18     return (father[x]);
19 }
20
21 int main () {
22     scanf("%d", &t);
23     for (int i = 1; i <= 30000; i++) {
24         father[i] = i;
25     }
26     for (int i = 1; i <= t; i++) {
27         scanf(" %c %d %d", &c, &x, &y);
28         if (c == 'M') {
29             int tx = getfather(x);
30             int ty = getfather(y);
31             if (tx != ty) {
32                 father[tx] = ty;
33                 dis[tx] = ++sum[ty];
34                 sum[ty] += sum[tx];
35             }
36         } else {
37             int tx = getfather(x);
38             int ty = getfather(y);
39             if (tx != ty) printf("-1\n");
40             else printf("%d\n", abs(dis[x] - dis[y]) - 1);
41         }
42     }
43
44     return 0;
45 }