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，
题目大意就是给定一棵树，要给他染色，每个格子染色需要一个单位的时间，每个格子有自己的权值，对于每个格子染色的消耗为权值和已经历的时间单位相乘，需要特别注意的是对一个格子进行染色时，他的上一个格子必须要染过色。
这道题贪心好想，代码化难，无非就是取一下权值进行排序，但是难点在于，他要求对一个格子进行染色时需保证他的父辈都染过色，一开始我是想外定时间t然后一点点的取权值最大做的，后发现关联性不够清晰，然后去网上借鉴了大佬的思路，开始了快乐找爹之旅。。。
简言之，就是每个格子都对着自己需要的染色次数，如果他是当前平均权值最大的那个，就把他和他的父辈并起来，如果父辈已被粉刷，就去找曾祖父。。以此类推就出现了快乐找爹之旅。
以下是代码
#include
#include
#includeusing namespace std;
const int N = 1005;
struct node {
int num_node; //此集合中包含点数
int father; //父亲
int sumc; //总权值
int ans; //它和它的子节点所用花费
int flag; //表示它已访问过 } Tree[N];int Find (int p)//并查集合并节点（快乐源泉） {
if (p != Tree[p].father && Tree[Tree[p].father].flag){
Tree[p].father = Find(Tree[p].father);
}
return Tree[p].father; } int main() {
int r,n,a,b,indexn,father,k;
double maxv;
while(scanf("%d%d",&n,&r), n || r ){
memset(Tree,0,sizeof(Tree));
for(int i=1; i<=n; i++){
scanf("%d",&Tree[i].sumc);
Tree[i].ans = Tree[i].sumc;
Tree[i].num_node = 1;
}
for(int i=1; i<n; i++){
scanf("%d%d",&a,&b);
Tree[b].father = a;
}
Tree[r].father = r;
for(int i=1; i<n; i++){
maxv = 0;
for(int j=1; j<=n; j++){
if(!Tree[j].flag && maxv < Tree[j].sumc1.0/Tree[j].num_node && j != r){
indexn = j;
maxv = Tree[j].sumc1.0/Tree[j].num_node;
}
}
Tree[indexn].flag = 1;
father = Find(indexn);
Tree[father].ans += Tree[indexn].ans + Tree[indexn].sumc * Tree[father].num_node;
Tree[father].sumc += Tree[indexn].sumc;
Tree[father].num_node += Tree[indexn].num_node;
}
printf("%d\n",Tree[r].ans);
}
return 0; }