POJ 1258 Agri-Net Prim/Kruskal

题意：将n个村庄连接起来，每两个村庄之间连接的消耗已经知道。求将所有村庄连在一起所需要的最小消耗。
题解：给出两组代码。
Prim:

#include<cstdio>
#define MAX 300
#define INF 99999999
int map[MAX][MAX];
int dis[MAX], vis[MAX];

int Prim ( int n )
{
    int i, j, k, m, sum = 0;
    for ( i = 0; i < n; i++ )
        dis[i] = INF, vis[i] = 0;
    sum = dis[0] = 0;
    for ( i = 0; i < n; i++ )
    {
        for ( m = INF, j = 0; j < n; j++ )
            if ( !vis[j] && dis[j] < m )
                m = dis[k=j];
        vis[k] = 1;
        sum += m;
        for ( j = 0; j < n; j++ )
            if ( !vis[j] && map[k][j] < dis[j] )
                dis[j] = map[k][j];
    }
    return sum;
}

int main()
{
    int n;
    while ( scanf("%d",&n) != EOF )
    {
        for ( int i = 0; i < n; i++ )
            for ( int j = 0; j < n; j++ )
                scanf("%d",&map[i][j]);
        printf("%d\n",Prim(n));
    }
    return 0;
}

 

Kruskal:

#include <iostream>
using namespace std;

int father[101], n;

struct edge
{
	int u,v,w;
} e[101*101/2];

int cmp(const void *a, const void *b)
{
	return (*(edge *)a).w - (*(edge *)b).w;
}

int find_set ( int x )
{
	if ( x != father[x] )
		father[x] = find_set(father[x]);
	return father[x];
}

int Kruskal ( int x )
{
	int i, sum = 0;
	for ( i = 0; i < n; i++ )
		father[i] = i;

	qsort(e,x,sizeof(edge),cmp);

	for ( i = 0; i < x; i++ )
	{
		int a = find_set(e[i].u);
		int b = find_set(e[i].v);
		if ( a != b )
		{
			father[b] = a;
			sum += e[i].w;
		}
	}
	return sum;
}

int main()
{
	while ( cin >> n )
	{
		int c = 0, temp;
		for ( int i = 0; i < n; i++ )
			for ( int j = 0; j < n; j++ )
			{
				cin >> temp;
				if ( i < j )
				{
					e[c].u = i;
					e[c].v = j;
					e[c].w = temp;
					c++;
				}
			}
			cout << Kruskal(c) << endl;
	}
	return 0;
}