Connect the Cities(MST prim)

Connect the Cities

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3371

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

 

 #include<stdio.h>
 #include<string.h>
 using namespace std;
 #define N 505
 #define INF 0x7fffffff
 int n , m , k ;
 int g[N][N] , vis[N] , low[N] ;
 
 void prim ()
 {
     int pos =  , res =  ;
     memset (vis ,  , sizeof(vis)) ;
     vis[pos] =  ;
     for (int i =  ; i <= n ; i++)
         if ( !vis[i] )
             low[i] = g[pos][i] ;
     for (int i =  ; i < n ; i++) {
         int minn = INF ;
         pos = - ;
         for (int j =  ; j <= n ; j++)
             if ( !vis[j] && low[j] < minn) {
                 minn = low[j] ;
                 pos = j ;
         }
         if (pos == -) {
             puts ("-1") ;
             return ;
         }
         res += minn ;
         vis[pos] =  ;
         for (int j =  ; j <= n ; j++)
             if (!vis[j] && low[j] > g[pos][j])
                 low[j] = g[pos][j] ;
     }
     printf ("%d\n" , res) ;
 }
 int main()
 {
    // freopen ("a.txt" , "r" , stdin);
     int u , v , w ;
     int T , t;
     scanf ("%d" , &T) ;
     while (T--) {
         scanf ("%d%d%d" , &n , &m , &k) ;
         for (int i =  ; i <= n ; i++)
             for (int j =  ; j <= n ; j++)
                 g[i][j] = INF ;
         for (int i =  ; i < m ; i++) {
             scanf ("%d%d%d" , &u , &v , &w) ;
             if (g[u][v] > w)//notice
                 g[u][v] = g[v][u] = w ;
         }
         for (int i =  ; i < k ; i++) {
             scanf ("%d%d" , &t , &u) ;
             for (int j = ; j <= t -  ; j++) {
                 scanf ("%d" , &v) ;
                 g[u][v] = g[v][u] =  ;
             }
         }
         prim () ;
     }
 }

给偶上了一课，本以为用kruskal就能一条路走到黑：

但遗憾的TLE了，

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，用prim较优

转载：http://www.cnblogs.com/GO-NO-1/p/3710816.html