POJ - 1287 Networking (最小生成树Kruskal)

题目:http://poj.org/problem?id=1287

题意:

P(<=50)个点,R(< long long)对边,找出最小通路和

分析:

长度为边权的最小生成树问题

注意边数的数目不限,注意long long,存入边数,和更新相同边的最小值

核心:

void Kruskal()
{
	memset(parent, -1, sizeof(parent));
	int num = 0, ans = 0, u, v;
	for(i = 0; i= n-1) break;
	}
	printf("%d\n", ans);
}
代码:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

#define MAXN 100
#define MAXM 100000
#define MIN -1e+10
#define INF 0x7f7f7f7f

struct edge
{
	int u, v, w;
	bool operator < (const edge& p)const
	{
		return w=0; s = parent[s]);
	while(s != x)
	{
		temp = parent[x];
		parent[x] = s;
		x = temp;
	}
	return s;
}

void Union(int u, int v)
{
	int r1 = Find(u), r2 = Find(v);
	int temp = parent[r1] + parent[r2];
	if(parent[r1] > parent[r2])
	{
		parent[r1] = r2;
		parent[r2] = temp;
	}
	else 
	{
		parent[r2] = r1;
		parent[r1] = temp;
	}
}

void Kruskal()
{
	memset(parent, -1, sizeof(parent));
	int num = 0, ans = 0, u, v;
	for(i = 0; i= n-1) break;
	}
	printf("%d\n", ans);
}

int vis[100][100];

int main()
{
	freopen("a.txt", "r", stdin);
	__int64 mt;
	while(~scanf("%d%I64d", &n, &mt) && n)
	{
		memset(vis, -1, sizeof(vis));
		int u, v, w;
		m = 0;
		while(mt--)
		{
			scanf("%d%d%d", &u, &v, &w);
			if(vis[u][v] == -1)
			{
				vis[u][v] = vis[v][u] = m;
				edges[m].u = u; edges[m].v = v; edges[m].w = w;
				m++;
			}
			else if(edges[vis[u][v]].w > w)
			{
				edges[vis[u][v]].w = w;
			}
		}
		sort(edges, edges+m);
		Kruskal();
	}
	return 0;
}



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