poj 2377 Bad Cowtractors

Bad Cowtractors

Description

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".

Input

* 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.

Output

* 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.

Sample Input

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

Sample Output

42

题意：给定每条路线间需要的费用，建造一个最贵的网络线路（任意两节点都可以互相达到，但是不存在回路），求最多需要花费多少钱。

思路：最大生成树。

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define M 1500000
int f[M];
int sum,t;
struct node
{
	int x,y,l;
}p[M],v[M];

bool cmp(node a,node b)
{
	return (a.l>b.l);
}

void Init (int n)
{
	for(int i=1;i<=n;i++)
		f[i]=i;
}

int find(int x)
{
	if(x!=f[x])
		return f[x]=find(f[x]);
	return f[x];
}

int Union(int x,int y,int z)
{
	int fx,fy;
	fx=find(x);
	fy=find(y);
	if(fx!=fy)
	{
		f[fy]=fx;
		sum+=z;
		t++;
	}
    return sum;
}

int main ()
{
	int n,m;
	int i;
	while(cin>>n>>m)
	{
		Init(n);
		memset(p,0,sizeof(p));
		sum=0;
		t=0;
		for(i=0;i<m;i++)
			cin>>p[i].x>>p[i].y>>p[i].l;

		sort(p,p+m,cmp);

		for(i=0;i<m;i++)
		  Union(p[i].x,p[i].y,p[i].l);

		if(t==n-1)
			cout<<sum<<endl;
		else
			cout<<"-1"<<endl;
	}
	return 0;
}

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