cf1042F. Leaf Sets(贪心)

题意

题目链接

给出一棵树，删除一些边，使得任意联通块内的任意点距离不超过$k$

sol

考场上想的贪心是对的：考虑一棵子树，如果该子树内最深的两个节点的距离相加$>k$就删掉最深的那个点，向上update的时候只返回最深的点的深度

然而却苦于写不出代码。。。

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define ll long long 
#define ull unsigned long long 
#define rg register 
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? eof : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *o = obuf;
//void print(int x) {if(x > 9) print(x / 10); *o++ = x % 10 + '0';}
//#define os  *o++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int maxn = 1e6 + 10, inf = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, k, ans = 0;
vector<int> v[maxn];
int dfs(int x, int fa) {
    if(v[x].size() == 1) return 0;
    vector<int> dis;
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(to == fa) continue;
        dis.push_back(dfs(to, x) + 1);        
    }
    sort(dis.begin(), dis.end());
    while(dis.size() >= 2) {
        int x = dis[dis.size() - 1], y = dis[dis.size() - 2];
        if(x + y > k) 
            ans++, dis.pop_back();
        else break;
    }
    return dis.back();
}
main() {
    n = read(); k = read();
    for(int i = 1; i <= n - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y);
        v[y].push_back(x);
    }
    for(int i = 1; i <= n; i++)
        if(v[i].size() > 1) {dfs(i, 0); break;}
    printf("%d", ans + 1);
    return 0;
}
/*
2 2 1
1 1
2 1 1
*/