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

#include <cstdio>
#include <cstring>
#include <algorithm>


const int maxn = 30000 + 500;
int father[maxn], dis[maxn];
int sum[maxn];
char c;
int x, y;
int t;

int getfather(int x) {
    if (father[x] == x) return (x);
    int cur = father[x];
    father[x] = getfather(father[x]);
    dis[x] = dis[x] + dis[cur];
    return (father[x]);
}

int main () {
    scanf("%d", &t);
    for (int i = 1; i <= 30000; i++) {
        father[i] = i;
    }
    for (int i = 1; i <= t; i++) {
        scanf(" %c %d %d", &c, &x, &y);
        if (c == 'M') {
            int tx = getfather(x);
            int ty = getfather(y);
            if (tx != ty) {
                father[tx] = ty;
                dis[tx] = ++sum[ty];
                sum[ty] += sum[tx];
            }
        } else {
            int tx = getfather(x);
            int ty = getfather(y);
            if (tx != ty) printf("-1
");
            else printf("%d
", abs(dis[x] - dis[y]) - 1);
        }
    }

    return 0;
}