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 = ;
 struct MFS {
     int fa[N], tm[N], cnt[N];
     void init() { for (int i = ; i < N; i++) fa[i] = i, tm[i] = , cnt[i] = ;}
     int find(int x, int &t) {
         if (fa[x] == x) {
             t = ;
         } 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] - ;
         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\n", fx, cnt[fx], tm[x] + tm[fx]);
     }
 } mfs;

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

——written by Lyon