hdu 3371(kruskal)
Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15569 Accepted Submission(s): 4108
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.
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.
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
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
int father[];
int dep[];
struct Seg
{
int x,y,len;
} seg[];
int cmp(Seg a,Seg b)
{
return a.len <b.len;
}
int _find(int x)
{
if(x==father[x])return x;
return _find(father[x]);
}
int main()
{
int tcase ;
scanf("%d",&tcase);
while(tcase--)
{
int n,m,t;
scanf("%d%d%d",&n,&m,&t);
for(int i=; i<=n; i++)
{
father[i]=i;
dep[i]=;
}
for(int i=; i<m; i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
seg[i].x=x;
seg[i].y=y;
seg[i].len=z;
}
for(int i=; i<t; i++)
{
int num,x;
scanf("%d%d",&num,&x);
num--;
while(num--)
{
int y;
scanf("%d",&y);
x = _find(x);
y = _find(y);
if(x!=y)
{
if(dep[x]==dep[y])
{
father[x] = y;
dep[y]++;
}
else if(dep[x]<dep[y])
{
father[x] = y;
}
else
{
father[y]=x;
} }
}
}
sort(seg,seg+m,cmp);
int sum=;
for(int i=; i<m; i++)
{
int x = _find(seg[i].x);
int y = _find(seg[i].y);
if(x!=y)
{
if(dep[x]==dep[y])
{
father[x] = y;
dep[y]++;
}
else if(dep[x]<dep[y])
{
father[x] = y;
}
else
{
father[y]=x;
}
sum+=seg[i].len;
}
}
int ans=;
for(int i=; i<=n; i++)
{
if(father[i]==i) ans++;
}
if(ans==) printf("%d\n",sum);
else printf("-1\n");
} return ;
}
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