843C - Upgrading Tree

843C - Upgrading Tree

题意：给一棵n个点的树，允许不超过2*n次操作。操作：（x,y,z），删除边（x,y），添加边（y,x），要求满足三个条件（见原题）

题解：

待补~

先贴上dalao代码：

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 const int N = 1234567;
 6
 7 vector <int> st;
 8 int pv[N], sz[N];
 9 vector <int> g[N];
10 vector <int> jobs[N];
11 int depth[N];
12
13 bool flag;
14
15 void dfs(int v, int pr) {
16   st.push_back(v);
17   pv[v] = pr;
18   sz[v] = 1;
19   for (int u : g[v]) {
20     if (u == pr) {
21       continue;
22     }
23     depth[u] = depth[v] + 1;
24     dfs(u, v);
25     sz[v] += sz[u];
26   }
27   if (flag && depth[v] >= 3) {
28     jobs[st[1]].push_back(v);
29   }
30   st.pop_back();
31 }
32
33 int main() {
34   int n;
35   scanf("%d", &n);
36   for (int i = 0; i < n - 1; i++) {
37     int x, y;
38     scanf("%d %d", &x, &y);
39     x--; y--;
40     g[x].push_back(y);
41     g[y].push_back(x);
42   }
43   flag = false;
44   dfs(0, -1);
45   vector <int> roots;
46   for (int i = 0; i < n; i++) {
47     int outer = n - sz[i];
48     if (outer > n / 2) {
49       continue;
50     }
51     bool ok = true;
52     for (int j : g[i]) {
53       if (j == pv[i]) {
54         continue;
55       }
56       if (sz[j] > n / 2) {
57         ok = false;
58         break;
59       }
60     }
61     if (ok) {
62       roots.push_back(i);
63     }
64   }
65   assert(!roots.empty());
66   flag = true;
67   if ((int) roots.size() == 1) {
68     depth[roots[0]] = 0;
69     dfs(roots[0], -1);
70   } else {
71     depth[roots[0]] = depth[roots[1]] = 0;
72     dfs(roots[0], roots[1]);
73     dfs(roots[1], roots[0]);
74   }
75   vector < pair <int, pair <int, int> > > res;
76   for (int i = 0; i < n; i++) {
77     if (jobs[i].empty()) {
78       continue;
79     }
80     int at = i;
81     for (int &v : jobs[i]) {
82       res.emplace_back(pv[i], make_pair(at, v));
83       res.emplace_back(v, make_pair(pv[v], i));
84       at = v;
85     }
86     res.emplace_back(pv[i], make_pair(at, i));
87   }
88   int cnt = res.size();
89   printf("%d\n", cnt);
90   for (int i = 0; i < cnt; i++) {
91     printf("%d %d %d\n", res[i].first + 1, res[i].second.first + 1, res[i].second.second + 1);
92   }
93   return 0;
94 }

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