hdu 3371 Connect the Cities (最小生成树)

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8718 Accepted Submission(s): 2437

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

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

Author

dandelion

Source

HDOJ Monthly Contest – 2010.04.04

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分析时间复杂度：

kruskal算法为O(e*lge)，prim算法为O(v*v)(貌似可用优先队列缩减)，此题数据为v=500，e=25000，理论上prim较优，但考虑到在kruskal算法find()处理过程中的路径压缩，使得两种算法耗时差不多。

 1 //546MS     1220K    1441B     C++    
 2 //G++ TLE，prim算法模板题 
 3 #include<stdio.h>
 4 #include<string.h>
 5 #define N 505
 6 #define INF 0x7ffffff
 7 int g[N][N];
 8 int low[N],vis[N];
 9 int n;
10 int prim()
11 {
12     int pos,res=0;
13     memset(vis,0,sizeof(vis));
14     vis[1]=1;
15     pos=1;
16     for(int i=1;i<=n;i++)
17         if(i!=pos) low[i]=g[pos][i];
18     for(int i=1;i<n;i++){
19         int minn=INF;
20         pos=-1;
21         for(int j=1;j<=n;j++)
22             if(!vis[j] && minn>low[j]){
23                 minn=low[j];
24                 pos=j;
25             }
26         if(pos==-1) return -1; //注意这里 
27         res+=minn;
28         vis[pos]=1;
29         //printf("**%d %d
",pos,minn);
30         for(int j=1;j<=n;j++)
31             if(!vis[j] && low[j]>g[pos][j])
32                 low[j]=g[pos][j];
33     }
34     return res;
35 }
36 int main(void)
37 {
38     int t,m,k;
39     int u,v,d;
40     scanf("%d",&t);
41     while(t--)
42     {
43         scanf("%d%d%d",&n,&m,&k);
44         for(int i=0;i<=n;i++) 
45             for(int j=0;j<=n;j++)
46                 g[i][j]=INF;
47         for(int i=0;i<m;i++){
48             scanf("%d%d%d",&u,&v,&d);
49             if(g[u][v]>d)    //notice 
50                 g[u][v]=g[v][u]=d; 
51         }
52         for(int i=0;i<k;i++){
53             scanf("%d",&d);
54             scanf("%d",&u);
55             d--;
56             while(d--){
57                 scanf("%d",&v);
58                 g[u][v]=g[v][u]=0;
59                 u=v;
60             }
61         }
62         printf("%d
",prim());
63     }
64     return 0;
65 }