HDU3371 最小生成树

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原文链接：http://www.cnblogs.com/--ZHIYUAN/p/6053395.html

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针对ConnecttheCities问题，采用Prim算法求解最小生成树。该问题描述了在若干已幸存城市间，通过构建道路实现所有城市再次连通的场景。输入包含城市数量、可选道路及费用、仍保持连接的城市组等信息。

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18067 Accepted Submission(s): 4460

Problem Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

Sample Input

1

   6 4 3
  

   1 4 2 
  

   2 6 1
  

   2 3 5
  

   3 4 33
  

   2 1 2 
  

   2 1 3 
  

   3 4 5 6
  

Sample Output

1

Author

dandelion

Source

HDOJ Monthly Contest – 2010.04.04

题意：

n个点，m条已知长度的边，k组已经连接的边，问最小要建多少条边；

代码：

 1 //prim 模板。这题数据太大用cruscal会超时。
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 int dis[502],map[502][502],mark[502],ha[502];
 6 const int MAX=10000007;
 7 int prim(int n)
 8 {
 9     for(int i=1;i<=n;i++)  //初始化每个点到生成树中点的距离
10     {
11         dis[i]=map[1][i];
12         mark[i]=0;
13     }
14     mark[1]=1; //1这个点加入生成树中。
15     int sum=0;
16     for(int i=1;i<n;i++) //枚举n-1条边
17     {
18         int sta=-1,Min=MAX;
19         for(int j=1;j<=n;j++)  //找不在生成树中的点中距离生成树中的点长度最小的
20         {
21             if(!mark[j]&&dis[j]<Min)
22             {
23                 Min=dis[j];
24                 sta=j;
25             }
26         }
27         if(sta==-1) return -1; //没找到可以可以联通的路
28         mark[sta]=1;   //新找到的点加入生成树
29         sum+=Min;       
30         for(int j=1;j<=n;j++)  //更新树外的点到树中的点的距离
31         {
32             if(!mark[j]&&dis[j]>map[sta][j])
33             dis[j]=map[sta][j];
34         }
35     }
36     return sum;
37 }
38 int main()
39 {
40     int n,m,k,p,q,c,h,t;
41     scanf("%d",&t);
42     while(t--)
43     {
44         scanf("%d%d%d",&n,&m,&k);
45         for(int i=0;i<=n;i++)
46         for(int j=0;j<=n;j++)
47         map[i][j]=MAX;
48         for(int i=0;i<m;i++)
49         {
50             scanf("%d%d%d",&p,&q,&c);
51             if(map[p][q]>c)        //可能有重边
52             map[p][q]=map[q][p]=c;
53         }
54         while(k--)
55         {
56             scanf("%d",&h);
57             for(int i=1;i<=h;i++)
58             {
59                 scanf("%d",&ha[i]);
60                 for(int j=1;j<i;j++)
61                 map[ha[i]][ha[j]]=map[ha[j]][ha[i]]=0;
62             }
63         }
64         int ans=prim(n);
65         printf("%d\n",ans);
66     }
67     return 0;
68 }

转载于:https://www.cnblogs.com/--ZHIYUAN/p/6053395.html