HDUOJ---3371Connect the Cities

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7997    Accepted Submission(s): 2267

Problem 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

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define maxn 25005
typedef struct Side
{
    int vs;
    int ve;
    int cost;
}side;
side sta[maxn];
int father[],rank[];
void init(int n)
{
    for(int i= ;i<n ; i++)
    {
        father[i]=i;
           rank[i]=;
    }
}

int setfind(int x)
{

    if(x!=father[x])
        father[x]=setfind(father[x]);
    return father[x];  //这里不能用x
}
void krusal(int a,int b)
{
    int x=setfind(a);
    int y=setfind(b);
    if(x==y) return ;
    if(x<y)
    {
     father[y]=x;
     rank[x]+=rank[y];
    }
    else
    {
    father[x]=y;
    rank[y]+=rank[x];
    }
}
int cmp(const void* a , const void* b)
{
    return (*(side*)a).cost-(*(side*)b).cost;
}
int main()
{
    int test,n,m,k,i,ans;
    scanf("%d",&test);
    while(test--)
    {
        ans=;
        scanf("%d%d%d",&n,&m,&k);
        init(n);
        for(i=; i<m ;i++)
        {
            scanf("%d%d%d",&sta[i].vs,&sta[i].ve,&sta[i].cost);
        }
        int a,b,c;
        for(i=;i<k ;i++)
        {
            scanf("%d",&c);
            scanf("%d",&a);
            while(c-- >)
            {
                scanf("%d",&b);
                krusal(a,b);
            }
        }
        qsort(sta,m,sizeof(sta[]),cmp);
            for(i=; i<m ;i++)
            {
                int tem1=setfind(sta[i].vs);
                int tem2=setfind(sta[i].ve);
                if(tem1 != tem2)
                {
                    krusal(tem1,tem2);
                    ans+=sta[i].cost;
                }
            }

        if(rank[]==n)
        printf("%d\n",ans);
        else
            printf("-1\n");
    }
    return ;
}