Constructing Roads

这题叙述的有点小问题，两个村庄无论经过几个其他的村庄只要相连即可。所有已这是最小生成树问题，已建成的距离设为0；

There 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.

Input The 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.

Output You 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 3 0 990 692 990 0 179 692 179 0 1 1 2

Sample Output 179

Source

#include
#include
using namespace std;




int dist[102][102];
struct edge{
  int i;
  int j;
  int value;
  
}p[10000];
bool cmp(edge a,edge b)
{
    return a.value

}
int parent[108];
void init(int n)
{
    for(int i=1;i<=n;i++)
{
 parent[i]=i;
}


}
int find(int i)
{
   if(i==parent[i])
   {
     return i;
   }
   parent[i]=find(parent[i]);
   return parent[i];
}
void combine(int a,int b)
{
    parent[a]=b;


}
int main()
{
//freopen("in.txt","r",stdin);
int n;
while(scanf("%d",&n)!=EOF)
{
  init(n);
  int i,j;
   int k=0;
  for(i=1;i<=n;i++)
  {
    for(j=1;j<=n;j++)
{
  scanf("%d",&dist[i][j]);
}
  }


  int m;
  scanf("%d",&m);
  int a,b;
  while(m--)
  {
  scanf("%d%d",&a,&b);
  dist[a][b]=0;
      dist[b][a]=0;
  }
      
  for(i=1;i<=n;i++)
  {
     for(j=i+1;j<=n;j++)
 {
      p[k].i=i;
  p[k].j=j;
  p[k].value=dist[i][j];
      k++;
 }
  }
  //cout<   sort(p,p+k,cmp);
  int res=0;
  for(i=0;i   {
      int c,d;
  c=find(p[i].i);
  d=find(p[i].j);
  if(c!=d)
  {
  //cout<<"c:"<   res+=p[i].value;
               combine(c,d);
     
  }
  
  }
  cout<  
}


  return 0;
}