poj 2377 Bad Cowtractors （最大生成树）

最新推荐文章于 2019-08-15 13:57:37 发布

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本文探讨了如何在给定的n个点和m条边中，寻找一个满足特定条件的最大生成树，即总成本最高，同时确保所有点相连且无环。文章通过详细解释算法流程，包括初始化、边的排序以及Kruskal算法的应用，展示了如何实现这一目标。

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题目链接：http://poj.org/problem?id=2377

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：n个点，m条边，求一个最大生成树
注意输出-1

最大生成树就是最小生成树中的排序函数改成从大到小排即可
#pragma GCC optimize(2)
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<set>
#include<vector>
#include<string>
#include<queue>
using namespace std;
const int maxn = 1500;
const int inf = 0x3f3f3f3f;
typedef long long ll;
struct node
{
	int u, v, w;
}edge[maxn*maxn];
int f[maxn];
int sum = 0;
int n, m;
bool cmp(node &a, node &b)
{
	return a.w > b.w;
}
int find(int x)
{
	if (x == f[x])
	{
		return x;
	}
	else
	{
		return f[x] = find(f[x]);
	}
}
int flag = 0;
void kru()
{
	int po = 0;
	for (int i = 0; i < m; i++)
	{
		int u = find(edge[i].u);
		int v = find(edge[i].v);
		if (u != v)
		{
			po++;
			sum += edge[i].w;
			f[u] = v;
			if (po == n - 1)
			{
				flag = 1;
				return;
			}
		}
	}
}
int main()
{
	//freopen("C://input.txt", "r", stdin);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
	{
		f[i] = i;
	}
	for (int i = 0;i < m; i++)
	{
		scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
	}
	sort(edge, edge + m, cmp);
	kru();
	if (flag)
	{
		printf("%d\n", sum);
	}
	else
	{
		printf("-1\n");
	}
	return 0;
}