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.
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.
#include<iostream>
#include<algorithm>
using namespace std;
int pre[100005];
struct node{
int a,b,c;
}r[100005];
bool cmp(node x,node y)
{
return x.c<y.c;
}
int find(int x)
{
return x==pre[x] ? x:pre[x]=find(pre[x]);
}
void join(int x,int y)
{
pre[find(y)]=find(x);
}
int main()
{
int n,q,cnt,temp,sum,c,a,b;
while(cin>>n)
{
cnt=0,sum=0,c=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
cin>>temp;
if(j<=i)
continue;
r[cnt].a=i;
r[cnt].b=j;
r[cnt].c=temp;
cnt++;
}
}
sort(r,r+cnt,cmp);
cin>>q;
for(int i=0;i<=n;i++)
{
pre[i]=i;
}
for(int i=0;i<q;i++)
{
cin>>a>>b;
int fa=find(a);
int fb=find(b);
if(fa!=fb)
{
c++;
pre[fb]=fa;
}
}
for(int i=0;i<cnt;i++)
{
int fa=find(r[i].a);
int fb=find(r[i].b);
if(fa!=fb)
{
sum+=r[i].c;
pre[fb]=fa;
c++;
}
if(c==n-1)
{
break;
}
}
cout<<sum<<endl;
}
}