11733 - Airports

[MRH]

I can not find my wrong in simple MST problem

Please give me some I/O

Thanks in advance

[robot]

Hi try this input checking.. its help u i think..
4
3 2 120
1 2 20
2 3 120
5 2 100
1 3 20
3 5 100
5 2 120
1 3 20
3 5 100
4 3 10
1 2 5
2 3 15
3 4 0
Case #1: 260 2
Case #2: 420 4
Case #3: 480 3
Case #4: 25 2
ASU(SUST)
"Teme jabo bole to Shopno deki ni"

[anis_cuet016]

Hi,

i m having difficulty in understanding the theme of this problem, at first i thought it can b solved by using kruskal and union find... but the sample i/o given below is not fitting to my logic...

5 2 100
1 3 20
3 5 100

here there are link between 1 -> 3 -> 5 so i consider it whole as 1, there are two more locations 2 and 4 so all together it became 1 + 1 + 1 = 3 airports...... but how the correct number of airports became 4 ??? please clarify me ....

sory 4 my bad English .....

[zobayer]

Well, indeed this can be solved by kruskal, but you have misunderstood the problem. Problem clearly said

The government now needs your help to decide on the cheapest way of ensuring that every location has access to an airport. The aim is to make airport access as easy as possible, so if there are several ways of getting the minimal cost, choose the one that has the most airports.

So, it you look closely, when you build the MST, you have larger edges at the end. Say, the last edge cost is A and airport building cost is B where A > B, so clearly, you would like not to build this road, rather, you will build an airport more. As you are putting just one airport in each tree of the minimum spanning forest, it is clear from the tree property that, if you remove an edge from it, it will be split into two trees.... So, previously you built one airport for the tree, and if the above property holds, you would cut this road, and then you have to build one more airport.

I think now you understand the problem.

[anis_cuet016]

Thankz a lot zobayer ......... what kind of shit i m, didnt give proper concentration in that para, i think i should b more carefull....... nwayz just added this line in my code and got accepted nthing else changed

Code: Select all

if(G[i].wt < weight)
// blah blah blah ......

Hey buddy thankz again, was stuck in this problem for more then two week ..........
Khodahafiz
Anis

[Shafaet_du]

try this:

Code: Select all

1
8 7 10
1 3 5
1 2 8
2 4 6
3 4 10
4 5 30
5 6 6
5 7 15

Code: Select all

Case #1: 65 4

[spewer]

Hi help me pls i cannot find the WA
tell pls if i misunderstood the problem or gimme a critical Input pls

Code: Select all

#include "stdafx.h"
#include<string>
#include<stdio.h>
#include<vector>
#include<string.h>
#include<utility>
#include<queue>

using namespace std;

vector<int> pset(100000);
int findSet(int i) {
   return (pset[i] == i) ? i :   
   (pset[i] = findSet(pset[i]));
}
void initSet(int _size) {
   pset.clear();   
   pset.resize(_size);
   for (int i=0; i<_size;i++) 
      pset[i]=i;
}
void unionSet(int i, int j) 
{
   pset[findSet(i)]=findSet(j);
}
bool isSameSet(int i, int j) {
   return findSet(i)==findSet(j);
}

typedef pair<int,pair<int,int> > iii;
priority_queue < iii , vector < iii > , greater < iii > > edgeList;
int kruskal(int n) { // n = #vertices, m= #aristas, retorna valor MST.
   int cont = 0; double mst_cost =  0;   initSet(n);
   while(!edgeList.empty() && cont < n-1)    {
      iii front = edgeList.top(); edgeList.pop();
      if ( !isSameSet(front.second.first,front.second.second) ) // nodo a y b
      {
         cont++; 
		 mst_cost += front.first; 
         unionSet(front.second.first,front.second.second);
      }
   }
   return mst_cost;
}


int casos,vertices,aristas,aero;
int ori,des,pes;
int main()
{
	freopen("in.txt","rt",stdin);
	freopen("out.txt","wt",stdout);
	scanf("%d",&casos);
	for(int c=0;c<casos;c++)
	{
		while(!edgeList.empty())
			edgeList.pop();
		scanf("%d %d %d",&vertices,&aristas,&aero);// aero= costo por creaeaeropuerto
		for(int a=0;a<aristas;a++)
		{
			iii coso;
			scanf("%d %d %d",&ori,&des,&pes);
			if(pes<aero)
			{
				coso.first=pes;coso.second.first=ori-1;coso.second.second=des-1;//ORIgen DEStino PESo
				edgeList.push(coso);
			}
		}
		int rpta=kruskal(vertices);
		for(int i=0;i<vertices;i++)
			findSet(i);
		int buliano=0;
		for(int i=0;i<vertices;i++)
		for(int j=0;j<i;j++)
		{
			if(pset[i]==pset[j])
			{
				buliano++;
				break;
			}
		}
		rpta+=(aero*(vertices-buliano));//vertices - buliano = cantidad de aeropuertos
		printf("Case #%d: %d %d\n",c+1,rpta,vertices-buliano);
	}
	return 0;
}

[brianfry713]

remove the lines:
#include "stdafx.h"
freopen("in.txt","rt",stdin);
freopen("out.txt","wt",stdout);

[SyFy]

OMG! Solved it with JAVA in 0.892.

But its really frustrating... had this code first:

Code: Select all

if(i != t - 1) 
out.printf("Case #%d: %d %d\n", i + 1, totalCost + (long)used.cardinality() * a, used.cardinality());
else out.printf("Case #%d: %d %d", i + 1, totalCost + (long)used.cardinality() * a, used.cardinality());

Got WAs...

changed to:

Code: Select all

//if(i != t - 1) 
out.printf("Case #%d: %d %d\n", i + 1, totalCost + (long)used.cardinality() * a, used.cardinality());
//else out.printf("Case #%d: %d %d", i + 1, totalCost + (long)used.cardinality() * a, used.cardinality());

Got AC...

but problem statement clearly says:

Your program should output exactly T lines, one for each case.

[brianfry713]

You should print a newline at the end of the last line on most problems.

[SyFy]

does it matter for instance if I will print 2 newlines ? and how can I know that when I am getting WAs in JAVA its not PE ?

You should print a newline at the end of the last line on most problems.

[brianfry713]

Yes it matters how many newlines you print. Don't count on ever getting PE. On a lot of problems extra or missing whitespace will give you WA.

[SyFy]

yep. but its very sad when you get WA and dont know whats wrong: algo or output.

Yes it matters how many newlines you print. Don't count on ever getting PE. On a lot of problems extra or missing whitespace will give you WA.

[Diaa.Abobaraka]

Gives me Time Limit ......any help please ?

Code: Select all

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Airports {
	static class Edge implements Comparable<Edge> {
		int firstNode, secondNode;
		int weight;

		public Edge(int firstNode1, int secondNode1, int d) {
			firstNode = firstNode1;
			secondNode = secondNode1;
			weight = d;
		}

		@Override
		public int compareTo(Edge edgeObject) {
			if (edgeObject.weight < weight) {
				return 1;
			} else if (edgeObject.weight == weight) {
				return 0;
			} else {
				return -1;
			}

		}
	}

	static int[] parentList;
	static ArrayList<Edge> edgeList;

	static int findParent(int i) {
		if (i == parentList[i])
			return i;
		return parentList[i] = findParent(parentList[i]);
	}

	static void connect(int i, int j) {
		parentList[findParent(i)] = parentList[findParent(j)];
	}

	static boolean isConnected(int i, int j) {
		if (findParent(i) == findParent(j))
			return true;
		return false;
	}

	public static void main(String[] args) throws NumberFormatException,
			IOException {
		StringBuilder sb = new StringBuilder();
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		int n, m, airportCost;
		int first, second, len;
		int sum, cnt;
		int tc = Integer.parseInt(bf.readLine());
		for (int t = 1; t <= tc; t++) {
			StringTokenizer st = new StringTokenizer(bf.readLine());
			n = Integer.parseInt(st.nextToken());
			m = Integer.parseInt(st.nextToken());
			airportCost = Integer.parseInt(st.nextToken());
			parentList = new int[n];
			for (int i = 0; i < n; i++)
				parentList[i] = i;
			edgeList = new ArrayList<>();
			int xx = m;
			for (int i = 0; i < xx; i++) {
				StringTokenizer st1 = new StringTokenizer(bf.readLine());
				first = Integer.parseInt(st1.nextToken());
				second = Integer.parseInt(st1.nextToken());
				len = Integer.parseInt(st1.nextToken());
				if (len < airportCost) {
					edgeList.add(new Edge(first - 1, second - 1, len));
				} else {
					m--;
				}

			}
			Collections.sort(edgeList);
			cnt = n;
			sum = 0;
			for (int i = 0; i < m && cnt > 1; i++) {
				first = edgeList.get(i).firstNode;
				second = edgeList.get(i).secondNode;
				len = edgeList.get(i).weight;
				if (!isConnected(first, second)) {
					cnt--;
					sum += len;
					connect(first, second);
				}
			}
			sum += cnt * airportCost;
			sb.append("Case #" + t + ": " + sum + " " + cnt + "\n");

		}
		System.out.println(sb);

	}

}

[lbv]

Gives me Time Limit ......any help please ?

This problem has a very tight time limit and a large amount of input, so a really efficient method to read data is needed. I'm not sure if StringTokenizer cuts it.

I used to write in Java for competitions a few years ago, so I went back and looked at the old I/O routines that I used, adapted them to your code, and got AC in 0.809s.

For reference, here's the relevant reader class:

Code: Select all

    class $R {
        InputStream in; byte b; byte[] buf; int bi, bz;
        $R(InputStream I) { in=I; buf=new byte[65536]; bi=bz=0; read(); }
        boolean hasNext() { skip(); return b >= 0; }
        void skip() { while (b >= 0 && b <= 32) read(); }
        String next() {
            StringBuilder sb = new StringBuilder();
            for (skip(); b > 32; read()) sb.append((char)b);
            return sb.length() == 0 ? null : sb.toString(); }
        String nextLine() {
            StringBuilder sb = new StringBuilder();
            for (; b != 10 && b != 13; read()) sb.append((char)b);
            while (b == 10 || b == 13) read();
            return sb.toString(); }
        int nextInt() {
            int ival = 0; boolean isgn = false;
            skip(); if (b == '-') { isgn = true; read(); }
            for (; b > 32; read()) ival = ival*10 + b-48;
            return isgn ? -ival : ival; }
        void read() {
            if (bi==bz) {
                bi=0; try { bz=in.read(buf); } catch(Exception e) { bz=-1; } }
            b = bz == -1 ? -1 : buf[bi++];  }
    }

Using this class goes something like this:

Code: Select all

$R myReader = new $R(System.in);

// whenever you need to read an integer
int foo = myReader.nextInt();

Also, notice that you should not print an extra blank line at the end of the output.