POJ 1258 Agri-Net（最小生成树）

Kruskal算法（运用并查集思想）：
#include<stdio.h>
#include<stdlib.h>
#define N 20010
int f[N], n;
struct node
{
    int s, e, d;
}no[N];
void Init()
{
    int i;
    for (i = 0; i <= n; i++)
        f[i] = i;
}
int cmp(const void *a, const void *b)
{
    node *s1 = (node *)a, *s2 = (node *)b;
    return s1->d - s2->d;
}
int Find(int x)
{
    if (f[x] != x)
        f[x] = Find(f[x]);
    return f[x];
}
int main ()
{
    int k, a, ans, i, j, ns, ne;
    while (scanf("%d", &n) != EOF)
    {
        ans = 0;
        k = 0;
        Init();
        for (i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
            {
                scanf("%d", &a);
                no[k].s = i;
                no[k].e = j;
                no[k].d = a;
                k++;
            }
        }
        qsort(no, k, sizeof(no[0]), cmp);
        for (i = 0; i < k; i++)
        {
            ns = Find(no[i].s);
            ne = Find(no[i].e);
            if (ns != ne)
            {
                f[ns] = ne;
                ans += no[i].d;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

Prim算法：
#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f
#define N 110
int Map[N][N], dist[N], visit[N], n, ans;
void Prim()
{
    int i, j, Min, idex;
    memset(visit, 0, sizeof(visit));
    visit[1] = 1;
    for (i = 1; i <= n; i++)
        dist[i] = Map[1][i];
    for (i = 1; i < n; i++)
    {
        Min = INF;
        for (j = 1; j <= n; j++)
        {
            if (dist[j] < Min && !visit[j])
            {
                Min = dist[j];
                idex = j;
            }
        }
        visit[idex] = 1;
        ans += Min; //统计最小权值
        for (j = 1; j <= n; j++)
        {
            if (!visit[j] && dist[j] > Map[idex][j])
                dist[j] = Map[idex][j];
        }
    }
}
int main ()
{
    int i, j;
    while (scanf("%d", &n) != EOF)
    {
        ans = 0;
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
                scanf("%d", &Map[i][j]);
        }
        Prim(); //假设每次将结点1先加入最小生成树，直到n个结点全部加入
        printf("%d\n", ans);
    }
    return 0;
}