题意：给定一棵n个结点的树，问：对于每个结点，能否通过删除一条边并添加一条边使得仍是树，并且删除该结点后得到的各个连通分量结点数 <= n/2？
题解：树形dp，两遍dfs，第一遍dfs求得以各个结点为根的子树的结点数，以及各个结点下面切掉某条边后最多可切出多少个结点；
第二遍dfs求得每个结点上面切掉某条边后最多可切出多少个结点。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define X first 4 #define Y second 5 typedef long long ll; 6 const int N = 4e5+1; 7 vector<int> ve[N]; 8 int num[N], maxson[N], down[N], up[N]; 9 int n; 10 void gmax(int& a, int b){ if(a < b) a = b;} 11 void dfs(int x, int fa){ 12 // printf("dfs x %d, fa %d ", x, fa); 13 num[x] = 1; 14 down[x] = maxson[x] = 0; 15 for(int i = 0; i < ve[x].size(); i++){ 16 int y = ve[x][i]; 17 if(y == fa) continue ; 18 dfs(y, x); 19 num[x] += num[y]; 20 gmax(down[x], num[y] <= n/2? num[y]: down[y]); 21 gmax(maxson[x], num[y]); 22 } 23 } 24 multiset<int>::iterator it; 25 void dfs2(int x, int fa){ 26 // printf("dfs2 x %d, fa %d ", x, fa); 27 multiset<int> se; 28 for(int i = 0; i < ve[x].size(); i++){ 29 int y = ve[x][i]; 30 if(y != fa) se.insert( num[y] <= n/2? num[y]:down[y] ); 31 } 32 33 for(int i = 0; i < ve[x].size(); i++){ 34 int y = ve[x][i]; 35 if(y != fa){ 36 if(n-num[y] <= n/2) 37 up[y] = n-num[y]; 38 else { 39 it = se.find( num[y] <= n/2? num[y]:down[y] ); 40 se.erase(it); 41 gmax(up[y], up[x]); 42 if(!se.empty()) 43 gmax(up[y], *se.rbegin()); 44 se.insert( num[y] <= n/2? num[y]:down[y] ); 45 } 46 dfs2(y, x); 47 } 48 } 49 } 50 51 int main(){ 52 int u, v; scanf("%d", &n); 53 for(int i = 1; i < n; i++){ 54 scanf("%d%d", &u, &v); 55 ve[u].push_back(v), ve[v].push_back(u); 56 } 57 dfs(1, -1); 58 dfs2(1, -1); 59 bool tag; 60 // for(int i = 1; i <= n; i++) 61 // cout << maxson[i] << ' ' << num[i] << ' ' << down[i] << ' ' << up[i] << endl; 62 for(int i = 1; i <= n; i++){ 63 if(maxson[i] <= n/2&&n-num[i] <= n/2)//不用切 64 tag = true; 65 else if(n-num[i] > n/2)//要切上面 66 tag = n-num[i]-up[i] <= n/2; 67 else//要切下面 68 tag = maxson[i]-down[i] <= n/2; 69 putchar(tag+'0'); 70 putchar(i == n? ' ':' '); 71 } 72 return 0; 73 }