POJ2377 Bad Cowtractors(最大生成树)

题意：

连接农场，要求输出最大权值和并且不能连成环，如果不能全部连接，输出-1

要点;

就是最小生成树变成最大罢了，Kruskal算法，排序改成从大到小，一次AC。

15347152

Seasonal

2377

Accepted

412K

63MS

C++

881B

2016-04-03 13:27:46

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#define maxn 100050
using namespace std;
int p[maxn];
int m, n;
struct edge
{
	int u, v, len;
}e[maxn];

bool cmp(edge a, edge b)
{
	return a.len > b.len;
}
void init()
{
	for (int i = 1; i <= m; i++)
		p[i] = i;
}
int find(int x)
{
	if (p[x] == x) return x;
	return p[x] = find(p[x]);
}
bool merge(int x, int y)
{
	x = find(x);
	y = find(y);
	if (x != y)
	{
		p[x] = y;
		return true;
	}
	return false;
}
int kruskal()
{
	init();
	sort(e, e + n, cmp);
	int sum = 0, edges = 0;
	for (int i = 0; i < n; i++)
	{
		if (merge(e[i].u, e[i].v))
		{
			sum += e[i].len;
			edges++;
		}
		if (edges + 1 == m)
			return sum;
	}
	return -1;
}

int main()
{
	while (~scanf("%d%d", &m, &n))
	{
		for (int i = 0; i < n; i++)
			scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].len);
		printf("%d\n", kruskal());
	}
	return 0;
}