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

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");elseprintf("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constpublic 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;elsereturn 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;  //sortArrays.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) ;      }  
}