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Head of a Gang

[编程题]Head of a Gang
  • 热度指数:5626 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 64M,其他语言128M
  • 算法知识视频讲解
One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made between the two persons. A "Gang" is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threthold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.

输入描述:
For each case, the first line contains two positive numbers N and K (both less than or equal to 1000), the number of phone calls and the weight threthold, respectively. Then N lines follow, each in the following format:
Name1 Name2 Time
where Name1 and Name2 are the names of people at the two ends of the call, and Time is the length of the call. A name is a string of three capital letters chosen from A-Z. A time length is a positive integer which is no more than 1000 minutes.


输出描述:
For each test case, first print in a line the total number of gangs. Then for each gang, print in a line the name of the head and the total number of the members. It is guaranteed that the head is unique for each gang. The output must be sorted according to the alphabetical order of the names of the heads.
示例1

输入

8 59
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10
8 70
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10

输出

2
AAA 3
GGG 3
0
来个java版的。DFS遍历图,找出所有联通子图,然后子图的边权总和。
package com.liyc.algs.pata;

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

/**
 * 构建Map<index, name>, 构建图g[name1][name2]
 * DFS遍历图,找出所有联通子图,然后子图的边权总和。
 */
public class A1034 {
	// 图
	private static int[][] g;
	// 顶点是否已访问过
	private static int[] v;
	// 顶点权重
	private static int[] w;
	// 阀值
	private static int threthold;
	// 总人数
	private static int personsN;
	public static void main(String[] args) throws IOException {
		BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
		String line1 = bf.readLine();
		String[] line1as = line1.split(" ");
		int totalcalls = Integer.valueOf(line1as[0]);
		threthold = Integer.valueOf(line1as[1]);
		int maxpersons = totalcalls * 2;
		int index = 0;
		Map<Integer, String> persons = new HashMap<>(maxpersons);
		g = new int[maxpersons][maxpersons];
		v = new int[maxpersons];
		w = new int[maxpersons];
		// 构建图
		for (int k = 0; k < totalcalls; k++) {
			String[] lines = bf.readLine().split(" ");
			String name1 = lines[0];
			String name2 = lines[1];
			int i = 0;
			if(!persons.containsValue(name1)){
				i = index++;
				persons.put(i, name1);
			} else{
				i = getKeyFromValue(persons, name1);
			}
			int j = 0;
			if(!persons.containsValue(name2)){
				j = index++;
				persons.put(j, name2);
			} else{
				j = getKeyFromValue(persons, name2);
			}
			int time = Integer.valueOf(lines[2]);
			g[i][j] += time;
			g[j][i] += time;
			w[j] = w[j] + time;
			w[i] = w[i] + time;
		}
		bf.close();
		personsN = persons.size();
		// 保存各个联通子图
		Set<Set<Integer>> subg = new HashSet<>();
		// 遍历图
		for (int i = 0; i < personsN; i++) {
			if( v[i] > 0){
				continue;
			}
			Set<Integer> set = new HashSet<>(maxpersons);
			subg.add(set);
			dfs(i, set);
		}
		int count = 0;
		Map<Integer, Integer> ps = new TreeMap<>(Comparator.comparing(persons::get));
		// 遍历各子图判断,存入ps
		for (Set<Integer> set : subg){
			if(set.size() <= 2){
				continue;
			}
			int max = -1;
			int sum = 0;
			for (int i : set) {
				int iw = w[i];
				sum += iw;
				if (max == -1 || iw > w[max]) {
					max = i;
				}
			}
			if( sum /2 <= threthold) {
				continue;
			}
			count++;
			ps.put(max, set.size());
		}
		System.out.println(count);
		if(count==0){
			return;
		}
		ps.forEach((k, v)->{
			String name = persons.get(k);
			System.out.println(name+" "+ps.get(k));
		});
	}

	private static int getKeyFromValue(Map<Integer, String> persons, String name1) {
		Set<Map.Entry<Integer, String>> set =  persons.entrySet();
		for (Map.Entry<Integer, String> en : set) {
			if(en.getValue().equals(name1)) {
				return en.getKey();
			}
		}
		return 0;
	}

	private static void dfs(int i, Set<Integer> list) {
		v[i] = 1;
		list.add(i);
		for (int j = 0; j < personsN; j++) {
			if(g[i][j] == 0){
				continue;
			}
			if(v[j] > 0) {
				continue;
			}
			dfs(j, list);
		}
	}
}




发表于 2019-10-18 16:56:40 回复(0)

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