Bad Cowtractors（最大生成树）

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

Hint:

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. 

题意：

给出N个点和 M条边，和每条边的花费。让你找到一个方案把每个点都连起来并且形成一棵树且花费最大。

思路：就用克鲁斯卡尔方法只不过从大的边开始枚举。

代码：

#include<stdio.h>
#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>
#include<vector>

#define INF 0x3f3f3f3f

using namespace std;

int N,M;
int map[1005][1005];
int pre[1005];
int sum;
struct D
{
	int a,b,c;
	D(int A,int B,int C)
	{
		a = A,b = B,c = C;
	}
};

struct cmp
{
	bool operator()(D a,D b)
	{
		return a.c<b.c;
	}
};

int find(int a)
{
	if(pre[a] == a)return a;
	return pre[a] = find(pre[a]);
}

int judge(int a,int b)
{
	int A = find(a);
	int B = find(b);
	if(A!=B)
	{
		pre[A] = B;
		return 1;
	}
	return 0;
}

priority_queue<D,vector<D>,cmp>Q;

int main()
{
	scanf("%d %d",&N,&M);
	for(int i=1 ; i<=N ; i++)pre[i] = i;

	while(M--)
	{
		int a,b,c;
		scanf("%d %d %d",&a,&b,&c);
		if(map[a][b]<c)map[a][b] = map[b][a] = c;//去重边。 
	}

	for(int i=1 ; i<=N ; i++)
	{
		for(int j=i+1; j<=N ; j++)
		{
			if(map[i][j]!=0)
			{
				Q.push(D(i,j,map[i][j]));//去过重边的结果入队 
			}
		}
	}
	int edge = 0;
	while(!Q.empty())
	{
		D mid = Q.top();
		Q.pop();
		if(judge(mid.a,mid.b))
		{
			sum += mid.c;
			edge++;
		}
		if(edge == N-1)break;
	}
	if(edge == N-1)printf("%d\n",sum);
	else printf("-1\n");
	return 0;
}