hdu 1055 Color a Tree （树+贪心）

Color a Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3080    Accepted Submission(s): 1110

Problem Description

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

 

贪心

 

推荐博客：https://blog.csdn.net/gatieme/article/details/49202739

 

#include
#include
#include<string.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int n,r;
struct node{
    double w;//当前到这点的代价 
    int c;//总权值（包括子节点） 
    int t;//时间（也是当前节点的所有节点节点个数（子节点+他自己）） 
    int pre;
}tree[1010];
int find_max(){//找最大 的节点，即到先达它能得到最大价值的那个 
    double Max=-1.0;
    int p;
    for(int i=1;i<=n;i++){
        if(Max//根节点不参与运算，因为在@处已经算了根结点直接到每个节点的花费 
            Max=tree[i].w;
            p=i;
        }
    }
    return p;
}
int main(){
    while(cin>>n>>r&&n+r){
        mem(tree,0);
        int ans=0;
        for(int i=1;i<=n;i++){
            cin>>tree[i].c;
            tree[i].w=tree[i].c;
            tree[i].t=1;
            ans+=tree[i].c;//先计算从根节点直接到达每个节点的值 
        }
        for(int i=1;i){
            int a,b;
            cin>>a>>b;
            tree[b].pre=a;
        }
        for(int i=1;i){
            int Max=find_max();
            int p=tree[Max].pre;
            tree[Max].w=0;//计算过将他赋值为0，以便后面的计算 
            ans+=tree[Max].c*tree[p].t;//从 Max的父节点到它所要的价值 
        //    cout<"<
            for(int i=1;i<=n;i++){
                if(tree[i].pre==Max){
                    tree[i].pre=p;//将原本Max的子节点改成p（即Max的父节点） 的子节点 
                }
            }
            tree[p].t+=tree[Max].t;//累加总时间 
            tree[p].c+=tree[Max].c;//累加总价值 
            tree[p].w=1.0*tree[p].c/tree[p].t;//!!!这里的当前价值是到达该点后所能得到的价值（平均值），这是贪心的核心 
        }
        cout<endl;
    }
    return 0; 
}

 

转载于:https://www.cnblogs.com/liuzuolin/p/10505473.html