hdu 3371 Connect the Cities（prim算法）

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=3371

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13628    Accepted Submission(s): 3679

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

给出城市的数量 n 以及 需要相联的城市及所需花销，再给出已经相联的城市。求将所有城市相联的最小花销。

对于已经相联的城市，我们也将他们视为没有相联，并且将花销置为0.这样就和普通的最小生成树一样了，从头开始寻找就可以了

【源代码】

#include <cstdio>
#include <algorithm>
#include <vector>
#include <iostream>
#define INF 0xfffffff
using namespace std;

int n,m,k;
const int maxn = 510;
struct node{
    int v,len;
    node(int v=0, int len = 0):v(v),len(len){}
};
vector <node> G[maxn];
int intree[maxn];
int minDist[maxn];
void init(){
    for(int i=0;i<maxn;i++){
        intree[i]=0;
        G[i].clear();
        minDist[i]=INF;
    }
}
int prim(int s){
    intree[s]=1;
    int ans=0;
    for(int i=0;i<G[s].size();i++){
        int vex = G[s][i].v;
        minDist[vex] = min(G[s][i].len,minDist[vex]);
    }
    for(int nodeNum=0;nodeNum<n-1;nodeNum++){
        int tmpMin=INF;
        int addNode;
        for(int i=1;i<=n;i++){ //从1 到 n 的城市标号 
            if(!intree[i]&&minDist[i]<tmpMin){
                tmpMin=minDist[i];
                addNode = i;
            }
        }
        if(tmpMin==INF) {
            // cout<<"bug"<<endl;
            return -1;
        }
        ans+=tmpMin;
        intree[addNode]=1;
        for(int i=0;i<G[addNode].size();i++){
            int vex = G[addNode][i].v;
            if(!intree[vex]&&G[addNode][i].len<minDist[vex])
                minDist[vex] = G[addNode][i].len;
        }
    }
    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d%d%d",&n,&m,&k);
        int v1,v2,len;
        int start=1;
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&v1,&v2,&len);
            G[v1].push_back(node(v2,len));
            G[v2].push_back(node(v1,len));
            if(i==0)
                start = v1;
        }
        for(int i=0;i<k;i++){
            int n;
            scanf("%d",&n);
            int num[110];
            for(int j=0;j<n;j++){
                scanf("%d",&num[j]);
            }
            for(int j=0;j<n;j++){
                for(int k=j+1;k<n;k++){
                    G[num[j]].push_back(node(num[k],0));
                    G[num[k]].push_back(node(num[j],0));
                }
            }
        }
        int ans=prim(start);
        printf("%d\n",ans);
    }
    return 0;
}