hdoj1863 畅通工程（Prime || Kruskal）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1863

思路

最小生成树问题，使用Prime算法或者Kruskal算法解决。这题在hdoj1233的基础上增加了判断是否能形成最小生成树的要求。

代码

Prime算法：

 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 using namespace std;

 const int INF = 0x7fffffff;
 const int N =  + ;
 int map[N][N];
 int dist[N];
 int n, m;

 void prime()
 {
     int min_edge, min_node;
     for (int i = ; i <= m; i++)
         dist[i] = INF;
     int ans = ;
     int now = ;
     for (int i = ; i < m; i++)
     {
         dist[now] = -;
         min_edge = INF;
         for (int j = ; j <= m; j++)
         {
             if (j != now && dist[j] >= )
             {
                 if (map[now][j] > )
                     dist[j] = min(dist[j], map[now][j]);
                 if (dist[j] < min_edge)
                 {
                     min_edge = dist[j];    //min_edge存储与当前结点相连的最短的边
                     min_node = j;
                 }
             }
         }
         if (min_edge == INF)    //不能形成最小生成树
         {
             puts("?");
             return;
         }
         ans += min_edge;
         now = min_node;
     }
     printf("%d\n", ans);
 }

 int main()
 {
     //freopen("hdoj1863.txt", "r", stdin);
     while (scanf("%d%d", &n, &m) ==  && n)
     {
         memset(map, , sizeof(map));
         int a, b, d;
         for (int i = ; i < n; i++)
         {
             scanf("%d%d%d", &a, &b, &d);
             map[a][b] = map[b][a] = d;
         }
         prime();
     }
     return ;
 }

Kruskal算法：

 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 using namespace std;

 struct Edge
 {
     int a, b, dist;

     Edge() {}
     Edge(int a, int b, int d):a(a),b(b),dist(d){}
     bool operator < (Edge edge)
     {
         return dist < edge.dist;
     }
 };

 const int N =  + ;
 vector<Edge> v;
 int p[N];
 int n, m;

 int find_root(int x)
 {
     if (p[x] == -)
         return x;
     else return find_root(p[x]);
 }

 void kruskal()
 {
     memset(p, -, sizeof(p));
     sort(v.begin(), v.end());    //将边按边长从短到长排序
     int ans = ;
     for (int i = ; i < v.size(); i++)
     {
         int ra = find_root(v[i].a);
         int rb = find_root(v[i].b);
         if (ra != rb)
         {
             ans += v[i].dist;
             p[ra] = rb;
         }
     }

     int cnt = ;
     for (int i = ; i <= m; i++)
         if (p[i] == -) cnt++;
     if (cnt > )    //连通分量个数多于1个，不能形成最小生成树
     {
         puts("?");
         return;
     }
     printf("%d\n", ans);
 }

 int main()
 {
     //freopen("hdoj1863.txt", "r", stdin);
     while (scanf("%d%d", &n, &m) ==  && n)
     {
         v.clear();
         int a, b, d;
         for (int i = ; i < n; i++)
         {
             scanf("%d%d%d", &a, &b, &d);
             v.push_back(Edge(a, b, d));
         }
         kruskal();
     }
     return ;
 }