杭電1102 Constructing Roads

標籤：杭電   1102   constructing roads   最小產生樹   

Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13518    Accepted Submission(s): 5128


Problem DescriptionThere are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

InputThe first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

OutputYou should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

30 990 692990 0 179692 179 011 2

Sample Output

179

 

這道題也是用的最小產生樹做的

代碼：

#include<stdio.h>
#include<string.h>
#define INF 1 << 30
int map[101][101] ;
int dis[101] ;
int used[101] ;
void prim( int N )
{ 
 for(int k = 1 ; k <= N ; k++)
 {
  dis[k] = map[1][k] ;
  used[k] = 0 ;
 }
 int sum = 0 ;
 for(int i = 1 ; i <= N ; i++ )
 {
  int min = INF ;
  int c = 0 ;
  for(int j = 1 ; j <= N ; j++ )
  {
   if(!used[j] && dis[j] < min )
   {
    min = dis[j] ;
    c = j ;
   }
  }
  used[c] = 1 ;
  for(j = 1 ; j <= N ; j++)
  {
   if(!used[j] && dis[j] > map[c][j])
    dis[j] = map[c][j] ;
  }
 }

 for(i = 1 ; i <= N ; i++)
  sum += dis[i] ;
 printf("%d\n", sum);
}
int main()
{
 int N = 0 ;
 while(~scanf("%d" , &N))
 {
     memset(map , 0 , sizeof(map) ) ;
  for(int i = 1 ; i <= N ; i++)
  {
   for(int j = 1 ; j <= N ; j++)
   {
    scanf("%d" , &map[i][j]) ;
   }
  }
  int Q = 0 ;
  scanf("%d" , &Q) ;
  int x = 0 , y = 0 ;
        for( int m = 1 ; m <= Q ; m++ )
  {
   scanf("%d%d" , &x , &y ) ;
   map[x][y] = map[y][x] = 0 ;//已經建好的樹不用再建了
  }
  prim( N ) ;
 }
 return 0 ;
}

 

杭電1102 Constructing Roads