Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

A couple of years ago, a new world wide crisis started, leaving many people with economical problems. Some workers of a particular company are trying to ask for an increase in their salaries.

The company has a strict hierarchy, in which each employee has exactly one direct boss, with the exception of the owner of the company that has no boss. Employees that are not bosses of any other employee are called workers. The rest of the employees and the owner are called bosses.

To ask for a salary increase, a worker should file a petition to his direct boss. Of course, each boss is encouraged to try to make their subordinates happy with their current income, making the company's profit as high as possible. However, when at least T percent of its direct subordinates have filed a petition, that boss will be pressured and have no choice but to file a petition himself to his own direct boss. Each boss files at most 1 petition to his own direct boss, regardless on how many of his subordinates filed him a petition. A boss only accounts his direct subordinates (the ones that filed him a petition and the ones that didn't) to calculate the pressure percentage.

Note that a boss can have both workers and bosses as direct subordinates at the same time. Such a boss may receive petitions from both kinds of employees, and each direct subordinate, regardless of its kind, will be accounted as 1 when checking the pressure percentage.

When a petition file gets all the way up to the owner of the company, all salaries are increased. The workers' union is desperately trying to make that happen, so they need to convince many workers to file a petition to their direct boss.

Given the company's hierarchy and the parameter T, you have to find out the minimum number of workers that have to file a petition in order to make the owner receive a petition.

Input 

There are several test cases. The input for each test case is given in exactly two lines. The first line contains two integers N and T ( 1N10⁵ , 1T100), separated by a single space. N indicates the number of employees of the company (not counting the owner) and T is the parameter described above. Each of the employees is identified by an integer between 1 and N. The owner is identified by the number 0. The second line contains a list of integers separated by single spaces. The integer B_i, at position i on this list (starting from 1), indicates the identification of the direct boss of employee i (0B_ii - 1).

The last test case is followed by a line containing two zeros separated by a single space.

Output 

For each test case output a single line containing a single integer with the minimum number of workers that need to file a petition in order to get the owner of the company to receive a petition.

Sample Input 

3 100 
0 0 0 
3 50 
0 0 0 
14 60 
0 0 1 1 2 2 2 5 7 5 7 5 7 5 
0 0

Sample Output 

3 
2 
5

---

題目描述：

給定長官-直接下屬關係，最底層員工為了加薪問題，打算發出聯署。
當長官的下屬中，超過 T% 的人發出請願，那麼會倍感壓力再向更上級長官發出請願，
最後傳到老闆那邊。

問最少聯署人數為多少。

題目解法：

dp[i] 表示 i 的所有下屬中，最少請願數量。

那麼 dp[i] = sigma(dp[j])

dp[j] 為前 T% 小的幾個。

#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> g[100005];
int n, t;
int dfs(int nd) {
    if(g[nd].size() == 0)   return 1;//leaf
    vector<int> petition;
    int i;
    for(i = 0; i < g[nd].size(); i++)
        petition.push_back(dfs(g[nd][i]));
    sort(petition.begin(), petition.end());
    int mn = petition.size()*t/100+(petition.size()*t%100 != 0);
    int sum = 0;
    for(i = 0; i < mn; i++)
        sum += petition[i];
    return sum;
}
int main() {
    int boss, i;
    while(scanf("%d %d", &n, &t) == 2 && n) {
        for(i = 0; i <= n; i++)
            g[i].clear();
        for(i = 1; i <= n; i++) {
            scanf("%d", &boss);
            g[boss].push_back(i);
        }
        printf("%d\n", dfs(0));
    }
    return 0;
}