UVaLive 3902 Network (无根树转有根树，贪心)

题意：一个树形网络，叶子是客户端，其他的是服务器。现在只有一台服务器提供服务，使得不超k的客户端流畅，但是其他的就不行了，

现在要在其他结点上安装服务器，使得所有的客户端都能流畅，问最少要几台。

析：首先这是一棵无根树，我们可以转成有根树，正好可以用原来的那台服务器当根，然后在不超过 k 的叶子结点，就可以不用考虑，

然后要想最少，那么就尽量让服务器覆盖尽量多的叶子，我们可以从最低的叶子结点开始向上查找最长的距离当服务器，这样是最优的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn], node[maxn];
int f[maxn];
bool cover[maxn];

void dfs(int u, int fa, int cnt){
    f[u] = fa;
    if(1 == G[u].size() && cnt > m)  node[cnt].push_back(u);
    for(int i = 0; i < G[u].size(); ++i) if(G[u][i] != fa)
        dfs(G[u][i], u, cnt+1);
}

void dfs2(int u, int fa, int cnt){
    cover[u] = true;
    if(cnt >= m)  return ;
    for(int i = 0; i < G[u].size(); ++i){
        int v = G[u][i];
        if(v == fa)  continue;
        dfs2(v, u, cnt+1);
    }
}

int solve(){
    memset(cover, false, sizeof cover);
    int ans = 0;
    for(int j = n-1; j > m; --j)
        for(int i = 0; i < node[j].size(); ++i){
            int v = node[j][i];
            if(cover[v])  continue;

            for(int k = 0; k < m; ++k) v = f[v];
            dfs2(v, -1, 0);
            ++ans;
        }
    return ans;
}

int main(){
    int T;  cin >> T;
    while(T--){
        int s;
        scanf("%d %d %d", &n, &s, &m);
        for(int i = 1; i <= n; ++i)  G[i].clear(), node[i].clear();
        for(int i = 1; i < n; ++i){
            int u, v;
            scanf("%d %d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        dfs(s, -1, 0);
        printf("%d\n", solve());
    }
    return 0;
}