題幹
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.
Input
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.
Output
For each test case, output a line containing the minimum total coloring cost required for Bob to color all the nodes.
Sample Input
5 1
1 2 1 2 4
1 2
1 3
2 4
3 5
0 0
Sample Output
33
題意，每個結點都有一個粉刷權值，第幾個訪問所消耗的代價就是權值乘以第幾次訪問！貪心，怎麼貪？經歷了覺得網站有問題以及換網站。
去CSDN去看了看大牛寫的博客，解題報告，不太明白。慢慢的摸索，抄代碼，修改，自己敲。比較 權值應該等於真實權值➗合併節點數，相當於這個節點由N個等權值結點組成。權值就是刷它之前所消耗的代價，這樣理解起來就不是很難。
這樣一來就是不斷從大到小歸併權值，直到root樹根。
便有了如下貪心準則：
1.要使代價小，必須儘早訪問權值較大的結點。
2.要訪問該結點，必須先訪問他的父節點。
3.訪問一個節結後，從該節點的父結點訪問該節點的子節點不需要 是消耗代價。
也就是說訪問了最大值的父節點就因該立刻訪問最大直結點。便可以找最大值節點開始訪問。
代碼如下
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<cstring>
using namespace std;
bool myfind();
struct object
{
int grade,fa,tim;
double weight;
}ob[1005];
int branch,root,flag,fa,kid,tem,mx=0;
int main()
{
ios::sync_with_stdio(false);
while(cin>>branch>>root)
{
tem=0;
memset(ob,0,sizeof(object)*1005);
if(branch==root&&branch==0) break;
for(int i=1;i<=branch;i++) cin>>ob[i].grade,ob[i].tim=1,ob[i].weight=ob[i].grade;
for(int i=1;i<branch;i++)
{
cin>>fa>>kid;
ob[kid].fa=fa;
}
while(myfind())
{
ob[ob[mx].fa].grade=ob[ob[mx].fa].grade+ob[mx].grade;
tem=tem+ob[ob[mx].fa].tim*ob[mx].grade;
ob[ob[mx].fa].tim=ob[ob[mx].fa].tim+ob[mx].tim;
//cout<<tem<<endl;
ob[mx].weight=0;
for(int i=1;i<=branch;i++)
{
if(ob[i].fa==mx) ob[i].fa=ob[mx].fa;
}
ob[ob[mx].fa].weight = 1.0*ob[ob[mx].fa].grade/ob[ob[mx].fa].tim ;
}
tem=tem+ob[root].grade;
cout<<tem+1<<endl;
}
return 0;
}
bool myfind()
{
double max=0;
flag=0;
for(int i=1;i<=branch;i++)
{
if(i==root) continue;
if(ob[i].weight>max)
{
max=ob[i].weight;
mx=i;
flag=1;
//cout<<i<<endl;
}
}
return flag;
}