hdu--6178(多校

　　题意：要在一棵 n 个点的树上放 k 只猴子，然后删掉尽量多的边，使得删边后，每只猴子都至少和另外一只猴子相连，问最后剩下的边数。

　　思路：其实dfs遍历一次看有多少个点-边-点就好了，比赛的时候就觉得要从树尾开始分，其实不是，dfs遍历，vis标记就好了。这题的输入很大，要用多校给过的读入挂。

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>
#include<queue>
#include<set>
#include<string>
#include<map>
#include <time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 100000+5;
const int N = 10;
const ll maxm = 1e7;
const int INF = 0x3f;
const int mod=1000;
const ll inf = 1e15 + 5;
const db eps = 1e-9;
namespace fastIO {
    #define BUF_SIZE 100000
    // fread -> read
    bool IOerror = 0;

    char nc() {
        static char buf[BUF_SIZE], *pl = buf + BUF_SIZE, *pr = buf + BUF_SIZE;
        if(pl == pr) {
            pl = buf;
            pr = buf + fread(buf, 1, BUF_SIZE, stdin);
            if(pr == pl) {
                IOerror = 1;
                return -1;
            }
        }
        return *pl++;
    }

    inline bool blank(char ch) {
        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
    }

    void read(int &x) {
        char ch;
        while(blank(ch = nc()));
        if(IOerror)
            return;
        for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }
    #undef BUF_SIZE
};
using namespace fastIO;
struct Edge{
    int u, v, next;
}e[maxn*2];
int head[maxn*2], cnt, vis[maxn*2];
int dan;

void init() {
    cnt=0, dan=0;
    memset(head, -1, sizeof(head));
    memset(vis, 0, sizeof(vis));
}
void add(int u, int v) {
    e[cnt].v=v, e[cnt].next=head[u];
    head[u]=cnt++;
}
void dfs(int u, int fa) {
    for (int i=head[u]; ~i; i=e[i].next) {
        int v=e[i].v;
        if (v!=fa && v!=u)  {
            dfs(v, u);
            if (!vis[v]) {
                if (!vis[u]) {
                    vis[u]=vis[v]=1;
                }
                else  dan++;
            }
        }
    }
}
void solve() {
    init();
    int n, k;
    read(n);  read(k);
    for (int i=1; i<=n-1; i++) {
        int u;  read(u);
        add(u, i+1);  add(i+1, u);
    }
    dfs(1, 0);
    if (!vis[1])  dan++;
    int can=(n-dan)/2;
    int ans=0;
    if (2*can<k)  {
        ans=k-can;
    }
    else  ans=(k+1)/2;
    printf("%d\n", ans);
}
int main() {
    int t = 1;
    //freopen("in.txt", "r", stdin);
    //scanf("%d", &t);
    read(t);
    while(t--)
        solve();
    return 0;
}