3371 Connect the Cities（最小生成树）

Connect the Cities

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

最小生成树板子题
AC代码：

#include<bits/stdc++.h>using namespace std;int pri[505][505];const int INF=0x3f3f3f;int n,m;int prim() {    int dis[505]={0};    for(int i=1;i<=n;i++) dis[i]=pri[1][i];    int book[505]={0};    book[1]=1;    int sum=0,min=0,k,i;    for( i=1;i<n;i++) {        min=INF;        //寻找最小的点且未被标记过        for(int j=1;j<=n;j++) {            if(book[j]==0&&dis[j]<min) {                min=dis[j];                k=j;            }        }        if(min==INF) return -1;        book[k]=1;        sum+=min;        //更新        for(int j=1;j<=n;j++)        {            if(book[j]==0&&dis[j]>pri[k][j])            dis[j]=pri[k][j];        }    }     return sum;}int main(){    ios::sync_with_stdio(0);    cin.tie(0);    int t;    cin>>t;    while(t--) {        int k;        cin>>n>>m>>k;        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                pri[i][j]=INF;        for(int i=0;i<=n;i++)pri[i][i]=0;        for(int i=0;i<m;i++) {            int u,v,w;            cin>>u>>v>>w;            if(pri[u][v]>w)pri[v][u]=pri[u][v]=w;        }        while(k--) {            int temp;            cin>>temp;            int ans[temp+2];            for(int i=0;i<temp;i++) cin>>ans[i];            for(int i=0;i<temp;i++)                for(int j=0;j<temp;j++)                pri[ans[i]][ans[j]]=0;        }        cout<<prim()<<endl;    }    return 0;}