hdu 3635 Dragon Balls (MFSet)

Problem - 3635

　　切切水题，并查集。

　　记录当前根树的结点个数，记录每个结点相对根结点的转移次数。1y~

代码如下：

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

using namespace std;
const int N = 11111;
struct MFS {
    int fa[N], tm[N], cnt[N];
    void init() { for (int i = 0; i < N; i++) fa[i] = i, tm[i] = 0, cnt[i] = 1;}
    int find(int x, int &t) {
        if (fa[x] == x) {
            t = 0;
        } else {
            fa[x] = find(fa[x], t);
            tm[x] += t;
            t = tm[x];
        }
        return fa[x];
    }
    void merge(int x, int y) {
        int tmp;
        int fx = find(x, tmp);
        int fy = find(y, tmp);
        tm[fy] -= tm[fx] - 1;
        cnt[fx] += cnt[fy];
        fa[fy] = fa[fx];
//        for (int i = 0; i < 5; i++) cout << fa[i] << ' '; cout << endl;
    }
    void query(int x) {
        int tmp;
        int fx = find(x, tmp);
        printf("%d %d %d
", fx, cnt[fx], tm[x] + tm[fx]);
    }
} mfs;

int main() {
    int T, n, m;
    scanf("%d", &T);
    for (int cas = 1; cas <= T; cas++) {
        scanf("%d%d", &n, &m);
        mfs.init();
        char op[2];
        int x, y;
        printf("Case %d:
", cas);
        for (int i = 0; i < m; i++) {
            scanf("%s", op);
            if (op[0] == 'T') {
                scanf("%d%d", &x, &y);
                mfs.merge(y, x);
            } else {
                scanf("%d", &x);
                mfs.query(x);
            }
        }
    }
    return 0;
}

——written by Lyon