首页 > 专题系列 > Java解POJ > POJ 2421 Constructing Roads [解题报告] Java
2013
11-11

POJ 2421 Constructing Roads [解题报告] Java

Constructing Roads

问题描述 :

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

输入:

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

输出:

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

样例输入:

3
0 990 692
990 0 179
692 179 0
1
1 2

样例输出:

179

解题代码:

//* @author: [email protected]
import java.util.Scanner;
public class Main
{
 static int[] bin;
 public static void main(String[] args)
 {
  Scanner in=new Scanner(System.in);
  int a=in.nextInt();
  int[][] p=new int[a][a];
  for(int i=0;i< a;i++)
   for(int j=0;j< a;j++)
    p[i][j]=in.nextInt();
  bin=new int[a];
  for(int i=0;i< a;i++)
	bin[i]=i;
  int n=in.nextInt();
  while((n--)!=0)
  {
   int w1=in.nextInt();
   int w2=in.nextInt();
   union(w1-1,w2-1);
  }
  int ans=0,min,tag,tag2;
  lv: while(true)
      {
	min=9999999;
	tag=-1;tag2=-1;
	boolean bb=false;
	for(int i=0;i< a-1;i++)
	{		
	  for(int j=i+1;j< a;j++)
	  {
	   if(root(j)!=root(i))
	   {
	    if(p[i][j]< min)
	     {
		min=p[i][j];
		tag=j;tag2=i;
	      }
	    bb=true;
	   }	
	  }
	if(!bb) break lv;	
	}
	union(tag2,tag);
	ans+=min;
      }
      System.out.println(ans);
    }
	
  static int  root(int b)
  {
   int a=b;
   while(bin[a]!=a)
	a=bin[a];
   bin[b]=a;
   return a;
  }

 static void union(int a,int b)
 {
	int a1=root(a);
	int b1=root(b);
	if(a1==b1) return;
	if(a1>b1) bin[a1]=b1;
	else bin[b1]=a1;
  }
}

  1. a是根先忽略掉,递归子树。剩下前缀bejkcfghid和后缀jkebfghicd,分拆的原则的是每个子树前缀和后缀的节点个数是一样的,根节点出现在前缀的第一个,后缀的最后一个。根节点b出现后缀的第四个位置,则第一部分为四个节点,前缀bejk,后缀jkeb,剩下的c出现在后缀的倒数第2个,就划分为cfghi和 fghic,第3部分就为c、c

  2. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better.