题解
我们将图中给出的$VOD$为根建立有根树，每次取出未覆盖的最深的节点第$k$级祖先，将其建为$VOD$，遍历一遍，将能够盖到的点标记。重复，直到盖完。贪心正确性可以证明。
1 //It is made by Awson on 2017.9.16 2 #include <map> 3 #include <set> 4 #include <cmath> 5 #include <ctime> 6 #include <queue> 7 #include <stack> 8 #include <cstdio> 9 #include <string> 10 #include <vector> 11 #include <cstdlib> 12 #include <cstring> 13 #include <iostream> 14 #include <algorithm> 15 #define LL long long 16 #define Max(a, b) ((a) > (b) ? (a) : (b)) 17 #define Min(a, b) ((a) < (b) ? (a) : (b)) 18 #define Abs(a) ((a) < 0 ? (-(a)) : (a)) 19 using namespace std; 20 const int N = 1000; 21 22 int n, s, k, u, v; 23 struct tt { 24 int to, next; 25 }edge[N*2+5]; 26 int path[N+5], top; 27 28 int dep[N+5], fa[N+5]; 29 bool isleave[N+5]; 30 31 struct node { 32 int u, dist; 33 node () { 34 } 35 node (int _u,int _dist) { 36 u = _u; dist = _dist; 37 } 38 bool operator < (const node & b) const{ 39 return dist < b.dist; 40 } 41 }; 42 priority_queue<node>Q; 43 44 int get_father(int u, int kth) { 45 if (!kth) return u; 46 return get_father(fa[u], kth-1); 47 } 48 void dfs2(int u, int kth) { 49 isleave[u] = false; 50 if (!kth) return; 51 for (int i = path[u]; i; i = edge[i].next) 52 dfs2(edge[i].to, kth-1); 53 } 54 void dfs(int u, int depth) { 55 dep[u] = depth; 56 for (int i = path[u]; i; i = edge[i].next) 57 if (dep[edge[i].to] == -1) { 58 dfs(edge[i].to, depth+1); 59 fa[edge[i].to] = u; 60 isleave[u] = false; 61 } 62 } 63 void add(int u, int v) { 64 edge[++top].to = v; 65 edge[top].next = path[u]; 66 path[u] = top; 67 } 68 void work() { 69 memset(path, 0, sizeof(path)); top = 0; 70 memset(dep, -1, sizeof(dep)); 71 memset(isleave, 1, sizeof(isleave)); 72 memset(fa, 0, sizeof(fa)); 73 scanf("%d%d%d", &n, &s, &k); 74 for (int i = 1; i < n; i++) { 75 scanf("%d%d", &u, &v); 76 add(u, v); add(v, u); 77 } 78 dfs(s, 0); 79 for (int i = 1; i <= n; i++) 80 if (isleave[i] && dep[i] > k) { 81 Q.push(node(i,dep[i])); 82 } 83 int ans = 0; 84 while (!Q.empty()) { 85 while (!isleave[Q.top().u]) { 86 Q.pop(); 87 if (Q.empty()) break; 88 } 89 if (Q.empty()) break; 90 int father = get_father(Q.top().u, k); 91 dfs2(father, k); 92 ans++; 93 } 94 printf("%d ", ans); 95 } 96 int main() { 97 int t; 98 scanf("%d", &t); 99 while (t--) 100 work(); 101 return 0; 102 }