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2013
11-11

POJ 2377 Bad Cowtractors [解题报告] Java

Bad Cowtractors

问题描述 :

Bessie has been hired to build a cheap internet network among Farmer John’s N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie.

Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a “tree”.

输入:

* Line 1: Two space-separated integers: N and M

* Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.

输出:

* Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.

样例输入:

5 8
1 2 3
1 3 7
2 3 10
2 4 4
2 5 8
3 4 6
3 5 2
4 5 17

样例输出:

42

温馨提示:

OUTPUT DETAILS:

The most expensive tree has cost 17 + 8 + 10 + 7 = 42. It uses the following connections: 4 to 5, 2 to 5, 2 to 3, and 1 to 3.

解题代码:

//* @author: Yeming Hu"[email protected]"
import java.util.*;
import java.io.*;

public class Main 
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(new BufferedInputStream(System.in));
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] mind = new int[n];
        boolean[] reached = new boolean[n];
        int[][] dist = new int[n][n];
        
        for(int i = 0; i < n; i++)
        {
            mind[i] = 0;
            reached[i] = false;
            
            for(int j = 0; j < n; j++)
            {
                dist[i][j] = 0;
            }
        }
        
        for(int i = 0; i < m; i++)
        {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();
            
            if(dist[a-1][b-1] < c)
            {
                dist[a-1][b-1] = c;
                dist[b-1][a-1] = c;
            }
        }
        
        boolean isPossible = true;
        int total = 0;
        reached[0] = true;
        for(int j = 0; j < n; j++)
        {
            if(!reached[j] && dist[j][0] > mind[j])
            {
                mind[j] = dist[j][0];
            }
        }
        for(int i = 1; i < n; i++)
        {
            int max = 0;
            int index = -1;
            for(int j = 0; j < n; j++)
            {
                if(!reached[j] && mind[j] > max)
                {
                    max = mind[j];
                    index = j;
                }
            }
            if(index == -1)
            {
                isPossible = false;
                break;
            }
            reached[index] = true;
            total += mind[index];
            for(int j = 0; j < n; j++)
            {
                if(!reached[j] && dist[j][index] > mind[j])
                {
                    mind[j] = dist[j][index];
                }
            }
        }
        if(isPossible)
        {
            System.out.println(total);
        }else
        {
            System.out.println(-1);
        }
    }
    
}

  1. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?

  2. int half(int *array,int len,int key)
    {
    int l=0,r=len;
    while(l<r)
    {
    int m=(l+r)>>1;
    if(key>array )l=m+1;
    else if(key<array )r=m;
    else return m;
    }
    return -1;
    }
    这种就能避免一些Bug
    l,m,r
    左边是l,m;右边就是m+1,r;