HDU3371--Connect the Cities（最小生成树）

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

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lcy

又是最小生成树的裸题

这是代码

#include<stdio.h>
#include<string.h>
#define N 505
#define INF 0x7ffffff
int g[N][N];
int low[N],vis[N];
int n;
int prim()
{
    int pos,res=0;
    memset(vis,0,sizeof(vis));
    vis[1]=1;
    pos=1;
    for(int i=1;i<=n;i++)
        if(i!=pos) low[i]=g[pos][i];
    for(int i=1;i<n;i++){
        int minn=INF;
        pos=-1;
        for(int j=1;j<=n;j++)
            if(!vis[j] && minn>low[j]){
                minn=low[j];
                pos=j;
            }
        if(pos==-1) return -1; //注意这里
        res+=minn;
        vis[pos]=1;
        //printf("**%d %d\n",pos,minn);
        for(int j=1;j<=n;j++)
            if(!vis[j] && low[j]>g[pos][j])
                low[j]=g[pos][j];
    }
    return res;
}
int main(void)
{
    int t,m,k;
    int u,v,d;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0;i<=n;i++)
            for(int j=0;j<=n;j++)
                g[i][j]=INF;
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&u,&v,&d);
            if(g[u][v]>d)    //notice
                g[u][v]=g[v][u]=d;
        }
        for(int i=0;i<k;i++){
            scanf("%d",&d);
            scanf("%d",&u);
            d--;
            while(d--){
                scanf("%d",&v);
                g[u][v]=g[v][u]=0;
                u=v;
            }
        }
        printf("%d\n",prim());
    }
    return 0;
}