题意:员工之间形成一棵树,上级可以给下级发奖金,任何一个人最多可以给一个下级发,并且发了奖金后就不能接受奖金。求总共最多可以产生多少的奖金流动

思路:每次选择没有下级并且有上级的员工a,令它的上级为b,那么让b给a发奖金,之后把a和b从树中删掉,这样处理直到不存在这样的员工a。也就是说每次让叶子员工接受奖金。简单证明:对于最优情况,叶子和它的兄弟集合还有它的父亲一定有一个接受了奖金,否则可以选叶子接受从父亲发的奖金,这样比原来增加了1个奖金;假设父亲接受了奖金或者兄弟接受了奖金,那么换成自己接收奖金,奖金数目不会变少,并且一定合法。所以这样选一定是最优的。

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#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; #define X first
#define Y second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define fillarray(a, b) memcpy(a, b, sizeof(a)) typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull; #ifndef ONLINE_JUDGE
namespace Debug {
void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
}
#endif // ONLINE_JUDGE template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ const int maxn = 5e5 + ; int son[maxn], fa[maxn]; int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
int n, x;
while (cin >> n) {
for (int i = ; i < n; i ++) {
scanf("%d", &x);
fa[i + ] = x;
son[x] ++;
}
queue<int> Q;
for (int i = ; i <= n; i ++) {
if (son[i] == ) Q.push(i);
}
vector<int> ans;
while (!Q.empty()) {
int H = Q.front(); Q.pop();
if (son[fa[H]] > ) {
ans.pb(H);
son[fa[H]] = ;
if (-- son[fa[fa[H]]] == ) Q.push(fa[fa[H]]);
}
}
cout << * ans.size() << endl;
sort(all(ans));
for (int i = ; i < ans.size(); i ++) {
printf("%d%c", ans[i], i == ans.size() - ? '\n' : ' ');
}
}
return ;
}
05-11 22:28