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;

}