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

 

Author

dandelion

 

Source

HDOJ Monthly Contest – 2010.04.04

 

#include <cstdio>

#include <cstring>

using namespace std;

const int maxn=505;

const int Max=0x3f3f3f3f;

int map[maxn][maxn],low[maxn],visit[maxn];

int n,m,k;

void prim()

{

    int i,j,pos,min,mst=0;

    memset(visit,0,sizeof(visit));

    pos=1;

    visit[1]=1;

    for(i=1;i<=n;i++)

        low[i]=map[pos][i];

    for(i=1;i<n;i++)

    {

        min=Max;

        for(j=1;j<=n;j++)/更新min的值

        {

            if(!visit[j] && min>low[j])

            {

                min=low[j];

                pos=j;

            }

        }

        mst+=min;

        if(mst>=Max) break;//说明这个图不连通

        visit[pos]=j;

        for(j=1;j<=n;j++)

        {

            if(!visit[j] && low[j]>map[pos][j])//更新low数组

                low[j]=map[pos][j];

        }

    }

    if(mst>=Max)

        printf("-1\n");

    else

        printf("%d\n",mst);

}

int main()

{

    int t,i,j,x;

    int p,q,c;

    int a[101];

    scanf("%d",&t);

    while(t--)

    {

        memset(map,Max,sizeof(map));

        scanf("%d%d%d",&n,&m,&k);

        for(i=0;i<m;i++)

        {

            scanf("%d%d%d",&p,&q,&c);

            if(map[p][q]>c)//这里也要注意一下，题目没说有重边，可是数据里面有，没加这个就wa

            map[p][q]=map[q][p]=c;

        }

        while(k--)//处理已连接的边，把权值赋0。

		{

			scanf("%d",&x);

			for(i=1;i<=x;i++)

				scanf("%d",&a[i]);

			for(i=1;i<=x;i++)

				for(j=1;j<=x;j++)

					map[a[i]][a[j]]=0;

		}

       prim();

    }

    return 0;

}