这道题还是比较简单的，对于当前节点，算出每个儿子需要的人数
然后再算出当前节点需要多少个人数，然后排个序加上去就好了。
#include<cstdio>
#include<vector>
#include<algorithm>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;const int MAXN = 112345;
vector<int> son[MAXN];
int T, n;int dp(int u)
{if(son[u].empty()) return 1;int k = son[u].size();vector<int> t;REP(i, 0, k)t.push_back(dp(son[u][i]));sort(t.begin(), t.end());int c = (k * T - 1) / 100 + 1, ans = 0;REP(i, 0, c)ans += t[i];return ans;
}int main()
{while(~scanf("%d%d", &n, &T) &&n){REP(i, 0, n + 1) son[i].clear();REP(i, 1, n + 1){int x;scanf("%d", &x);son[x].push_back(i);}printf("%d\n", dp(0));}return 0;
}