hdu3371 Connect the Cities

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 868    Accepted Submission(s): 308

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

 

Recommend

lcy

// First use BFS or Union-Find Set to rebuild the map, then use minimal spanning tree.

#include<iostream> #include<vector> #include<algorithm> #define maxn 505 using namespace std; struct edge { int u,v,w;//起点,终点,权 edge(int _u,int _v,int _w):u(_u),v(_v),w(_w){//构造函数初始化 } bool operator<(const edge& e)const{//重载<运算符,实现排序 return w<e.w;//非降序 } }; vector<edge> g;//边集 int n,m,k;//顶点数,边数,已连接点 struct unionset//并查集 { int p[maxn];// void clear(){//初始化 for(int i=0;i<n;++i) p[i]=i; } int find(int x){//查找 return x==p[x]?x:p[x]=find(p[x]); } void UNION(int x,int y){//合并 p[find(x)]=find(y); } }set; void init()//初始化 { g.clear(); scanf("%d%d%d",&n,&m,&k); for(int i=0,u,v,w;i<m;++i){//m条边 scanf("%d%d%d",&u,&v,&w); g.push_back(edge(--u,--v,w)); } } void slove() { set.clear(); for(int i=0,t,s,p;i<k;++i){//已连接点 scanf("%d%d",&t,&s); --s; for(int j=1;j<t;++j){ scanf("%d",&p); set.UNION(s,--p); } } sort(g.begin(),g.end()); int ans=0; for(vector<edge>::iterator it=g.begin();it!=g.end();++it){//Kruskal最小生成树 int x=set.find(it->u),y=set.find(it->v); if(x!=y){ ans+=it->w; set.UNION(x,y); } } int root=set.find(0); for(int i=1;i<n;++i)//判断未连接点 if(set.find(i)!=root){ puts("-1"); return; } printf("%d/n",ans); } int main() { int t; scanf("%d",&t); while(t--){ init(); slove(); } }