POJ1251Jungle Roads(最小生成树)

题目大意：首先给你一个图，需要你求出最小生成树，输入N个节点，用大写字母表示了节点，然后节点与节点之间有权值。
比如有9个节点，然后接下来有n-1行表示了边的情况，拿第一行举例
A 2 B 12 I 25

表示A有两个邻点，B和I，AB权值是12，AI权值是25

邻接矩阵prime:

#include
#include
#include
using namespace std;
const int maxn = 30;
const int inf = 0x3f3f3f3f;
int road[maxn][maxn], dis[maxn];
bool vis[maxn];
int n;
void prim()
{
    int mi, v;
    for(int i = 0; i < n; i++)
    {
        dis[i] = road[0][i];
        vis[i] = false;
    }
    for(int i = 1; i <= n; i++)//包括第一个点在内，一共要纳入n个点
    {
        mi = inf;
        for(int j = 0; j < n; j++)//每次找出未纳入顶点集与已知顶点集构成的权值最小的一条边
        {
            if(!vis[j] && mi > dis[j])
            {
                v = j;
                mi = dis[j];
            }
        }
        vis[v] = true;//把该顶点纳入已知集合
        for(int j = 0; j < n; j++)//更新与未纳入集合中的顶点的边的最小权值
        {
            if(!vis[j] && dis[j] > road[v][j])
                dis[j] = road[v][j];
        }
    }
    for(int i = 1; i < n; i++)
        dis[0] += dis[i];
    cout << dis[0] << endl;
}
int main()
{
    int m, w;
    char s[5], en[5];
    while(cin >> n, n)
    {
        memset(road, 63, sizeof(road));
        for(int i = 0; i < n; i++)
            road[i][i] = 0;
        for(int i = 1; i < n; i++)
        {
            scanf("%s%d", s, &m);
            for(int j = 0; j < m; j++)
            {
                scanf("%s%d", en, &w);
                road[s[0]-'A'][en[0]-'A'] = road[en[0]-'A'][s[0]-'A'] = w;
            }
        }
        prim();
    }
    return 0;
}

kruskal:

#include
#include
using namespace std;
int p[27];
struct edge
{
    int villagea;
    int villageb;
    int distance;
}roads[80];
bool cmp(edge a, edge b)
{
    return a.distance < b.distance;
}
int find_ancestor(int x)
{
    return  p[x] == x ? x : p[x] = find_ancestor(p[x]);
}
int main()
{
    int n, k, distance, number, ans;
    char start_village, end_village;
    while(cin >> n, n)
    {
        ans = k = 0;
        for(int i = 0; i < 27; i++)
            p[i] = i;
        for(int i = 0; i < n- 1; i++)
        {
            cin >> start_village >> number;
            for(int j = 0; j < number; j++, k++)
            {
                cin >> end_village >> distance;
                roads[k].villagea = start_village - 'A';
                roads[k].villageb = end_village - 'A';
                roads[k].distance = distance;
            }
        }
        sort(roads, roads + k, cmp);
        for(int i = 0; i < k; i++)
        {
            int x = find_ancestor(roads[i].villagea);
            int y = find_ancestor(roads[i].villageb);

            if(x != y)
            {
                ans = ans + roads[i].distance;
                p[y] = x;
            }
        }
        cout << ans << endl;
    }
    return 0;
}