POJ2054.Color a Tree

最新推荐文章于 2020-03-24 10:57:18 发布

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#POJ #贪心 #Greedy

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本文探讨了在一个特殊树形结构中寻找最小染色成本的问题。该问题要求按特定顺序染色节点，使得最终的成本最小。文章提供了一个详细的算法实现，并通过一个示例解释了如何确定最优染色顺序。

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试题请参见: http://poj.org/problem?id=2054
Also available on: http://acm.hdu.edu.cn/showproblem.php?pid=1055

题目概述

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.

Problem Image

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.

解题思路

(过几个小时来填坑)

源代码

import java.util.Scanner;

public class Main {
    private class Node {
        public Node(int cost) {
            this.cost = cost;
            parentIndex = -1;
        }

        public int cost;
        public int parentIndex;
    }

    public int getMinCost(Node[] nodes, int rootIndex) {
        int n = nodes.length, totalCost = 0;
        boolean[] isVisited = new boolean[n];
        int[] totalCosts = new int[n];
        int[] totalNodes = new int[n];

        for ( int i = 0; i < n; ++ i ) {
            totalCosts[i] = nodes[i].cost;
            totalNodes[i] = 1;
        }
        for ( int i = 1; i < n; ++ i ) {
            int k = getBestNodeIndex(isVisited, totalCosts, totalNodes, rootIndex);
            int p = nodes[k].parentIndex;
            while ( isVisited[p] ) {
                p = nodes[p].parentIndex;
            }

            isVisited[k] = true;
            totalCost += totalCosts[k] * totalNodes[p];

            totalCosts[p] += totalCosts[k];
            totalNodes[p] += totalNodes[k];
        } 
        return totalCost + totalCosts[rootIndex];
    }

    private int getBestNodeIndex(boolean[] isVisited, 
        int[] totalCosts, int[] totalNodes, int rootIndex) {
        int bestIndex = 0;
        double bestValue = 0;

        for ( int i = 0; i < isVisited.length; ++ i ) {
            if ( isVisited[i] || i == rootIndex ) {
                continue;
            }

            double currentValue = (double)totalCosts[i] / totalNodes[i];
            if ( currentValue > bestValue ) {
                bestValue = currentValue;
                bestIndex = i;
            }
        }
        return bestIndex;
    }

    public static void main(String[] args) {
        Main s = new Main();
        Scanner in = new Scanner(System.in);

        while ( in.hasNext() ) {
            int n = in.nextInt(), r = in.nextInt();
            Node[] nodes = new Node[n];

            if ( n == 0 && r == 0 ) {
                break;
            }
            for ( int i = 0; i < n; ++ i ) {
                int cost = in.nextInt();
                nodes[i] = s.new Node(cost);
            }
            for ( int i = 1; i < n; ++ i ) {
                int v1 = in.nextInt(), v2 = in.nextInt();
                nodes[v2 - 1].parentIndex = v1 - 1;
            }
            System.out.println(s.getMinCost(nodes, r - 1));
        }
        in.close();
    }
}