Question

我使用了不相交的集合概念来解决这个问题。从访问过的所有挤奶员开始，每当有新分支到达新人时，他被标记为已访问。我已经测试了以下边缘情况：

市内没有挤奶员
每个人都是送奶工
有些送奶工没有与任何人联系
某人与任何人无关
2人之间的多条路径
Milkmen和人正在形成不相交的集合

我还在接受WA。关于遗失案件的任何想法？

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

/**
 * Created by VIVEK VERMA on 9/25/2016.
 */
public class SpojIITWPC4I {
    public static void main(String[] arge) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(reader.readLine());
        for (int t = 0; t < T; t++) {
            StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
            int n = Integer.parseInt(tokenizer.nextToken());
            int m = Integer.parseInt(tokenizer.nextToken());
            int totalIncludedNodes = 0;
            ManNode[] people = new ManNode[n];
            Queue<ManLink> links = new PriorityQueue<>();
            tokenizer = new StringTokenizer(reader.readLine());
            for(int i=0;i<n;i++){
                int val=Integer.parseInt(tokenizer.nextToken());
                people[i] = new ManNode(i);
                if(val==1){
                    people[i].isIncluded = true;
                    totalIncludedNodes++;
                }
            }

            for(int i = 0; i<m;i++){
                tokenizer = new StringTokenizer(reader.readLine());
                int start = Integer.parseInt(tokenizer.nextToken()) -1;
                int destination = Integer.parseInt(tokenizer.nextToken()) -1;
                int value = Integer.parseInt(tokenizer.nextToken());
                ManLink link = new ManLink(people[start], people[destination], value);
                people[start].AddChild(link);
                people[destination].AddChild(link);
                if(people[start].isIncluded ^ people[destination].isIncluded){
                    links.add(link);
                }
            }

            int minCost = 0;
            while(!links.isEmpty() && totalIncludedNodes <n){
                ManLink link = links.poll();
                ManNode nextAddedNode = (link.source.isIncluded==true?link.destinationNode:link.source);
                if(nextAddedNode.isIncluded){
                    continue;
                }
                minCost += link.value;
                nextAddedNode.isIncluded = true;
                totalIncludedNodes++;
                for(int i=0;i<nextAddedNode.children.size();i++){
                    if((nextAddedNode.isIncluded ^ nextAddedNode.children.get(i).source.isIncluded)
                            || (nextAddedNode.isIncluded ^ nextAddedNode.children.get(i).destinationNode.isIncluded)){
                        links.add(nextAddedNode.children.get(i));
                    }
                }
            }

            if(totalIncludedNodes == n){
                System.out.println(minCost);
            }else{
                System.out.println("impossible");
            }
        }
    }
}


class ManNode{
    boolean isIncluded = false;
    List<ManLink> children = new ArrayList<>();
    int index;

    ManNode(int index){
        this.index = index;
    }

    void AddChild(ManLink link){
        children.add(link);
    }
}

class ManLink implements Comparable{
    ManNode destinationNode, source;
    int value;
    ManLink(ManNode source, ManNode node, int value){
        this.source = source;
        this.destinationNode = node;
        this.value = value;
    }

    @Override
    public int compareTo(Object o) {
        return this.value - ((ManLink)o).value;
    }
}

获取WA（错误的asnwer）http://www.spoj.com/problems/IITWPC4I/

0 个答案: