hdoj3371Connect the Cities

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 13720 Accepted Submission(s): 3710

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

1prim：
#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f
#define max 500+10
int visit[max],map[max][max],low[max],used[max];
int city,road,connect,n;
void prim()
{
    int i,j,next;
    int min,mincost=0;
    for(i=1;i<=city;i++)
    {
        visit[i]=0;
        low[i]=map[1][i];
    } 
    visit[1]=1;
    for(i=2;i<=city;i++)
    {
        min=INF;
        next=1;
        for(j=1;j<=city;j++) //寻找距离最近的点 
        {
            if(!visit[j]&&min>low[j])
            {
                min=low[j];
                next=j;
            }
        }
        if(min==INF)
        {
            printf("-1\n");
            return ;
        }
        visit[next]=1;
        mincost+=min;
        for(j=1;j<=city;j++)//更新路径 
        {
            if(!visit[j]&&map[next][j]<low[j])
            low[j]=map[next][j];
        }
    }
    printf("%d\n",mincost);
}
int main()
{
    int t,i,j,x,y,c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&city,&road,&connect);
        for(i=1;i<=city;i++)
        {
            for(j=1;j<=city;j++)
            {
                if(i==j)
                map[i][j]=0;
                else
                map[i][j]=INF;
            }
        }
        while(road--)
        {
            scanf("%d%d%d",&x,&y,&c);
            if(map[x][y]>c)
            map[x][y]=map[y][x]=c;
        }
        while(connect--)
        {
            scanf("%d",&n);
            for(i=0;i<n;i++)
            scanf("%d",&used[i]);
            for(i=0;i<n;i++)
            {
                for(j=i+1;j<n;j++)
                {
                    map[used[i]][used[j]]=map[used[j]][used[i]]=0;
                }
            }
        }
        prime();
    }
    return 0;
}