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
题意：给出一个树以及他的节点及其权值，现在要把这个树上的所有节点染色，染色规则是根节点可以随意染色，但是对于其他节点要染色它的父节点必须染色，每次染色的代价为T*A[i],其中T代表了当前是第几次染色，求最小的染色总代价
思路：贪心 利用等效权值（该点包含的原始权值总和/该点包含的原始点数），不断选取等效权值最大的点p，然后与其父节点fa结合，合并之前p与fa各自的染色顺序是已知的，让p中的第一个节点接着fa的最后一个节点染色，直到整个树合并成一个节点，这个点存贮的顺序即为节点顺序
#include<iostream>
#include <cstdio>
using namespace std;
const int max_n=1010;
int pre[max_n];//pre[i]用来表示当前集合（包含i）的上个元素
int nex[max_n];//next[i]用来表示当前集合（包含i）的下个元素
int c[max_n];
int num[max_n];//num[i]表示当前集合（包含i）的元素个数
int visit[max_n];
int sum[max_n];//当前集合元素和
int father[max_n];//father[i]表示i的父元素
int n,r;
int find_max()
{ //找到当前权值最大的集合，合并后的点当做一个集合，也就是一个点
double imax=0;
int bh=-1;
for(int i=1; i<=n; i++) {
if(imax<(sum[i]*1.0)/num[i]&&visit[i]==0) {
imax=(sum[i]*1.0)/num[i];
bh=i;
}
}
return bh;
}
void make(int x)
{ //联合
int i;
for(i = father[x]; pre[i] != -1; i = pre[i]);//找到父元素所在的集合
sum[i]+=sum[x];
num[i]+=num[x];
for(i= father[x]; nex[i] != -1; i = nex[i]);//找到父元素所在集合的底元素然后合并
nex[i]=x;
pre[x]=i;
visit[x]=1;
}
int main()
{
while(scanf("%d %d",&n,&r),n&&r) {
for(int i=1; i<=n; i++) { //初始化
scanf("%d",&c[i]);
sum[i]=c[i];
visit[i] =0;
pre[i] = nex[i] = -1;
num[i] = 1;
}
for(int i = 1; i < n; i++) {
int a,b;
scanf("%d %d",&a,&b);
father[b] = a; //父子关系
}
int d;
visit[r]=1;
while(1) {
d=find_max();
if(d==-1)break;
make(d);
}
int ans=0,cnt=1;
for(int i=r; i!=-1; i=nex[i]) {
ans+=cnt*c[i];
cnt++;
}
printf("%d\n",ans);
}
return 0;
}