poj 2054:Color a Tree 貪心

2054:Color a Tree

總時間限制: 

1000ms 

內存限制: 

65536kB

描述

Bob is very interested in the data structure of a tree. A tree is a directed graph in which a special node is singled out, called the "root" of the tree, and there is a unique path from the root to each of the other nodes.

Bob intends to color all the nodes of a tree with a pen. A tree has N nodes, these nodes are numbered 1, 2, ..., N. Suppose coloring a node takes 1 unit of time, and after finishing coloring one node, he is allowed to color another. Additionally, he is allowed to color a node only when its father node has been colored. Obviously, Bob is only allowed to color the root in the first try.

Each node has a "coloring cost factor", Ci. The coloring cost of each node depends both on Ci and the time at which Bob finishes the coloring of this node. At the beginning, the time is set to 0. If the finishing time of coloring node i is Fi, then the coloring cost of node i is Ci * Fi.

For example, a tree with five nodes is shown in Figure-1. The coloring cost factors of each node are 1, 2, 1, 2 and 4. Bob can color the tree in the order 1, 3, 5, 2, 4, with the minimum total coloring cost of 33.

Given a tree and the coloring cost factor of each node, please help Bob to find the minimum possible total coloring cost for coloring all the nodes.

輸入

The input consists of several test cases. The first line of each case contains two integers N and R (1 <= N <= 1000, 1 <= R <= N), where N is the number of nodes in the tree and R is the node number of the root node. The second line contains N integers, the i-th of which is Ci (1 <= Ci <= 500), the coloring cost factor of node i. Each of the next N-1 lines contains two space-separated node numbers V1 and V2, which are the endpoints of an edge in the tree, denoting that V1 is the father node of V2. No edge will be listed twice, and all edges will be listed. 

A test case of N = 0 and R = 0 indicates the end of input, and should not be processed.

輸出

For each test case, output a line containing the minimum total coloring cost required for Bob to color all the nodes.

樣例輸入

5 1
1 2 1 2 4
1 2
1 3
2 4
3 5
0 0

樣例輸出

33

來源

Beijing 2004

#include<iostream>//基本思想：除了根結點外，任何一個價值最大的節點的父節點被訪問之後必須馬上訪問這個節點。
#include<cmath>//於是可以把這個點和父節點合併成一個節點，並修改權值，不斷重複直到只剩一個節點。 
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
using namespace std;
struct Node
{
	int father;
	int c;
	double w;//記錄權值，反映平均代價 
	int cnt;//記錄當前這個節點是由幾個節點合併成的 
}N[1005];
int n,r;
int main()
{
	while(cin>>n>>r)
	{
		if(n==0&&r==0)break;
		int sum=0;
		for(int i=1;i<=n;++i)
		{
			cin>>N[i].c;
			N[i].w=N[i].c;
			N[i].cnt=1;
			sum+=N[i].c;
		}
		for(int i=0;i<n-1;++i)
		{
			int s,t;
			cin>>s>>t;
			N[t].father=s;
		}
		for(int i=1;i<=n-1;++i)
		{
			int pos=0;//找權值最大的節點 
			double Max=-1;
			for(int j=1;j<=n;++j)
			{
				if(N[j].w>Max&&j!=r)
				{
					Max=N[j].w;
					pos=j;
				}
			}
			N[pos].w=0;
			int f=N[pos].father;
			sum+=N[pos].c*N[f].cnt;
			for(int j=1;j<=n;++j)//把權值最大的節點的子節點的父節點修改 
			{
				if(N[j].father==pos)
				{
					N[j].father=f;
				}
			}
			N[f].c+=N[pos].c;
			N[f].cnt+=N[pos].cnt;
			N[f].w=(double)N[f].c/N[f].cnt;//記錄平均代價 
		}
		cout<<sum<<endl;
	}
	return 0;
}