hdu 3371 Connect the Cities (Prim)

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

题意：已知若干路径和已经有联系的城市，求最小花费使所有城市相连

注意：必须使用c语言的输入输出，否则会超时

#include <iostream>#include <memory.h>#include <stdio.h>#define INF 0x3f3f3f3f#define NUM 505using namespace std;int n;bool vis[NUM];int lowcost[NUM];int g[NUM][NUM];int Prim(){    int ans=0;    memset(vis,false,sizeof(vis));    memset(lowcost,INF,sizeof(lowcost));    for(int i=1;i<=n;i++)        lowcost[i]=g[1][i];    vis[1]=true;    lowcost[1]=0;    for(int i=2;i<=n;i++)    {        int Min=INF;        int k=-1;        for(int j=1;j<=n;j++)        {            if(!vis[j]&&lowcost[j]<Min)            {                Min=lowcost[j];                k=j;            }        }        if(Min==INF)            return -1;        ans+=Min;        vis[k]=true;        for(int j=1;j<=n;j++)        {            if(!vis[j]&&lowcost[j]>g[k][j]&&g[k][j]!=INF)            {                lowcost[j]=g[k][j];            }        }    }    return ans;}int main(){    int T;    scanf("%d",&T);    int m,k,t,a,b;    int p,q,c;    while(T--)    {        scanf("%d%d%d",&n,&m,&k);        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                g[i][j]=INF;            }        }        while(m--)        {            scanf("%d%d%d",&p,&q,&c);            if(c<g[p][q])            g[p][q]=g[q][p]=c;        }        while(k--)        {            scanf("%d%d",&t,&a);            t=t-1;            while(t--)            {                scanf("%d",&b);                g[a][b]=g[b][a]=0;            }        }        int ans=Prim();        printf("%d\n",ans);    }    return 0;}