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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=0;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?1:-1; 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?1:-1;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 + 7; 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 = 1; i < n; i ++) { scanf("%d", &x); fa[i + 1] = x; son[x] ++; } queue<int> Q; for (int i = 1; i <= n; i ++) { if (son[i] == 0) Q.push(i); } vector<int> ans; while (!Q.empty()) { int H = Q.front(); Q.pop(); if (son[fa[H]] > 0) { ans.pb(H); son[fa[H]] = 0; if (-- son[fa[fa[H]]] == 0) Q.push(fa[fa[H]]); } } cout << 1000 * ans.size() << endl; sort(all(ans)); for (int i = 0; i < ans.size(); i ++) { printf("%d%c", ans[i], i == ans.size() - 1? ‘\n‘ : ‘ ‘); } } return 0; } |
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原文地址:http://www.cnblogs.com/jklongint/p/4761066.html