HDU 1301 Jungle Roads

这一题是比较水的最小生成树的题，作为模板题比较好，下面给出了prim算法跟kruskal算法的源代码。

AC code：

//kruskal
#include <iostream>
#include <algorithm>
using namespace std;
const int MAX = 30;
int father[MAX];
int n, sum;
int roadNumber;
struct ROAD{
    int s, e, v;
}road[MAX * MAX];


void init()     //对并查集初始化
{
    for (int i = 1; i <= n; i++)
    {
         father[i] = i;
    } 
}
int cmp(ROAD a, ROAD b)
{
    return a.v < b.v;
}
int find(int k)      //使用并查集判断是否产生了回路
{
    return k == father[k] ? k : father[k] = find(father[k]);
}
void kruskal()
{
    int i;
    for(i = 0; i < roadNumber; i++)
    {
         int s = find(road[i].s);
         int e = find(road[i].e);
         if (s != e)
         {
             sum += road[i].v;
             father[s] = e;
         }
    }
}
int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        int i, j, temp, tv;
        char a, b;
        roadNumber = 0, sum = 0;
        init();
        for(i = 0; i < n - 1; i++)
        {
            cin >> a >> temp;
            for(j = 0; j < temp; j++)
            {
                cin >> b >> tv;
                road[roadNumber].s = a - 'A' + 1;
                road[roadNumber].e = b - 'A' + 1;
                road[roadNumber++].v = tv;
            }
        }
        sort(road, road + roadNumber, cmp);    //对所有边按升序排序
        kruskal();
        cout << sum << endl;
    }
    return 0;
}

//prim
#include <iostream>
using namespace std;
const int MAX = 30;
const int INF = 100000;
int closedge[MAX];
int map[MAX][MAX];
int n, vis[MAX], sum;

void prim()
{
     int i, j;
     for(i = 1; i <= n; i++){
         closedge[i] = INF;
     }
     memset(vis, 0, sizeof(vis));
     closedge[1] = 0;
     for(i = 1; i <= n; i++){
        int min = INF;
        int pos = -1;
        for(j = 1; j <= n; j++){          //找出未加入生成树的点到已生成的最小生成树的最短距离
            if(!vis[j] && closedge[j] < min){
                min = closedge[j];
                pos = j;
            }
        }
        vis[pos] = 1;
        sum += min;
        for(j = 1; j <= n; j++){             //松弛操作，将到新加入的点距离最小的value赋给closedge
            if(map[pos][j] < closedge[j] && !vis[j])
                closedge[j] = map[pos][j];
        }
     }
}
int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        int i, j;
        char s, e;
        int temp, tv, ts, te;
        sum = 0;
        memset(map, 127, sizeof(map));
        for(i = 0; i < n - 1; i++){
              cin >> s >> temp;
              ts = s - 'A' + 1;
              for(j = 0; j < temp; j++){
                cin >> e >> tv;
                te = e - 'A' + 1;
                map[ts][te] = map[te][ts] = tv;
              }
        }
        prim();
        cout << sum << endl;
    }
    return 0;
}