HDU 3371

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7040    Accepted Submission(s): 2008

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

16 4 31 4 22 6 12 3 53 4 332 1 22 1 33 4 5 6

 

 

Sample Output

1

 

 

Author

dandelion

 

 

Source

HDOJ Monthly Contest – 2010.04.04

 

 

Recommend

lcy

 

 

题意是说有n个城市，其中有m条边未建，然后还有已经建好的。让你求最少费用。

一开始用prime做超时，以为算法不行，便用Kruskal做了一遍也是超时。最后才知道原来要用C++交，而不是G++

 

#include<stdio.h>#include<string.h>#define max 10000000int map[505][505];int lowcost[505];int visit[505];int n;int prime(){    int i,j;    int min,pos,sum=0;    pos=1;    memset(visit,0,sizeof(visit));    for(i=1; i<=n; i++)    {        lowcost[i]=map[1][i];    }    visit[1]=1;    for( i=1; i<n; i++)    {        min=max;        for( j=2; j<=n; j++)j=1也可            if(!visit[j]&&min>lowcost[j])            {                min=lowcost[j];                pos=j;            }            if(min==max)            {                return -1;            }            visit[pos]=1;            sum=sum+min;            for( j=2; j<=n; j++)j=1也可            {                if(!visit[j]&&lowcost[j]>map[pos][j])                    lowcost[j]=map[pos][j];            }    }    return sum;}int main(){    int te;    int m,k;    int i,j;    int p,q,c;    int t;    int temp[505];    scanf("%d",&te);        while(te--)        {            scanf("%d%d%d",&n,&m,&k);            for(i=1;i<=n;i++)            {                for(j=1;j<=n;j++)                {                    map[i][j]=max;                    map[i][i]=0;                }            }            while(m--)            {                scanf("%d%d%d",&p,&q,&c);                if(map[p][q]>c)                map[p][q]=map[q][p]=c;            }            while(k--)            {              scanf("%d",&t);              for(i=0;i<t;i++)              {                  scanf("%d",&temp[i]);              }              for(i=0;i<t;i++)              {                  for(j=i+1;j<t;j++)                  {                      map[temp[i]][temp[j]]=map[temp[j]][temp[i]]=0;                  }              }              }         int result=prime();         if(result==-1)         printf("-1\n");         else         printf("%d\n",result);}}

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