PKU/POJ 2378 Tree Cutting

预处理出每个子树的大小，就可以线性判断每个点拿掉之后剩余的子树是否符合条件。

1 #include <cstdio>
2 #include <algorithm>
3 #include <vector>
4 using namespace std;
5
6 const int N = 10000 + 10;
7
8 int n, half;
9 vector<int> adj
, ans;
10 int size
;
11
12 int cal_size(int u, int f)
13 {
14 size[u] = 1;
15 for (vector<int>::iterator it = adj[u].begin(); it != adj[u].end(); ++it)
16 {
17 if (*it != f)
18 size[u] += cal_size(*it, u);
19 }
20 return size[u];
21 }
22
23 void dfs(int u, int f)
24 {
25 bool flag = size[0] - size[u] <= half;
26 for (vector<int>::iterator it = adj[u].begin(); it != adj[u].end(); ++it)
27 {
28 if (*it != f)
29 {
30 if (size[*it] > half)
31 flag = false;
32 dfs(*it, u);
33 }
34 }
35 if (flag)
36 ans.push_back(u + 1);
37 }
38
39 int main()
40 {
41 scanf("%d", &n);
42 half = n / 2;
43 for (int i = 1; i < n; ++i)
44 {
45 int u, v;
46 scanf("%d%d", &u, &v);
47 --u,--v;
48 adj[u].push_back(v);
49 adj[v].push_back(u);
50 }
51 cal_size(0, -1);
52 dfs(0, -1);
53 sort(ans.begin(), ans.end());
54 for (vector<int>::iterator it = ans.begin(); it != ans.end(); ++it)
55 printf("%d\n", *it);
56 return 0;
57 }
58