POJ 2075 Tangled in Cables (c++/java)

标签：acm   编程   poj   mst   

http://poj.org/problem?id=2075

题目大意：

给你一些人名，然后给你n条连接这些人名所拥有的房子的路，求用最小的代价求连接这些房子的花费是否满足要求。

思路：

昨天20分钟的题，输入不小心写错了- -|||||看世界杯半场休息随便看了下发现了。。。。T T

用map进行下标的映射，然后求MST即可。

c++

#include<cstdio>
#include<string>
#include<map>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN = 500;
int fa[MAXN];
struct edge
{
	int from, to;
	double val;
	bool operator < (const edge& x)const
	{
		return val < x.val;
	}
}e[MAXN*MAXN];
map<string, int> m;

int find(int cur)
{
	return cur == fa[cur] ? cur : fa[cur] = find(fa[cur]);
}

int main()
{
	int len = 0, n;
	double a;
	cin >> a >> n;
	while (n--)
	{
		string temp;
		cin >> temp;
		m[temp] = len++;
	}
	cin >> n;

	for (len = 0; len<n; len++)
	{
		string from, to;
		double value;
		cin >> from >> to >> value;
		e[len].from = m[from];
		e[len].to = m[to];
		e[len].val = value;
	}

	for (int i = 0; i<len; i++)
		fa[i] = i;
	sort(e, e + len);

	double ans = 0;
	for (int i = 0; i<len; i++)
	{
		int from = e[i].from;
		int to = e[i].to;
		int root_x = find(from);
		int root_y = find(to);
		if (root_x == root_y) continue;

		fa[root_x] = root_y;
		ans += e[i].val;
	}
	if (ans >  a)
		printf("Not enough cable\n");
	else
		printf("Need %.1lf miles of cable\n", ans);
	return 0;
}

JAVA:

import java.math.BigDecimal;
import java.text.DecimalFormat;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {

	//final 相当于const
	public static final int MAXN=500;
	//写起来好不习惯
	public static int[] fa=new int[MAXN];
	public static TreeMap<String, Integer> m=new TreeMap<String, Integer>();
	public static edge[] e=new edge[MAXN*MAXN];
	public static int find(int cur)
	{
		//不能这么写？
		//return cur == fa[cur] ? cur : fa[cur] = find(fa[cur]);  
		if(cur==fa[cur])
			return cur;
		else
			return fa[cur] = find(fa[cur]);	
	}  
	
	
	
	public static void main(String[] args) {
		 int len = 0, n;  
		 double a;  
		
		 Scanner in=new Scanner(System.in);
		 a=in.nextDouble();
		 n=in.nextInt();
		 
		 while((n--)!=0)
		 {
			 String temp=in.next();
			 m.put(temp, new Integer(len++));		 
		 }
		 
		 n=in.nextInt();
		
		 double value;  
		 for (len = 0; len<n; len++)  
		 {  
		       String from=in.next();
		       String to=in.next();
		        value=in.nextDouble();
		        e[len]=new edge();
		        e[len].from = m.get(from);  
		        e[len].to = m.get(to);  
		        e[len].val = value;  
		  }  
		 
		 for (int i = 0; i<len; i++)  
		        fa[i] = i;  
		 
		//sort
		 Arrays.sort(e,0,len);   
		 double ans=0;
		 for(int i=0;i<len;i++)
		 {
			  int from = e[i].from;  
		      int to = e[i].to;  
		      int root_x = find(from);  
		      int root_y = find(to);  
		      if (root_x == root_y) continue;
		      
		      fa[root_x] = root_y;  
		      ans += e[i].val;  
		 }
			 
		  if (ans >  a)  
			  System.out.print("Not enough cable\n"); 
		    else   
		      System.out.printf("Need %.1f miles of cable\n", ans);	
		  //java 是.1f
	}

}

class edge implements  Comparable<edge>  
{
	int from,to;
	double val;
	public int compareTo(edge x)  
    {          
		//double比较错了一次)
		BigDecimal data1 = new BigDecimal(this.val); 
		BigDecimal data2 = new BigDecimal(x.val); 
		
        return data1.compareTo(data2) ;      
    }  
}

POJ 2075 Tangled in Cables (c++/java),布布扣,bubuko.com

POJ 2075 Tangled in Cables (c++/java)

标签：acm   编程   poj   mst   

原文地址：http://blog.csdn.net/murmured/article/details/37647347