NYOJ 38 布线问题

#include <stdio.h>
#include <string.h>

int e[501][501];//储存地图 
int h[501];

int mintree;
int v,l;


int prime()//普里姆算法 
{
    int i,j;
    int book[501];//判断回路用的，book为1的表示已经加入树节点的楼 
    int temp;
    int count=0;//表示加入树节点的数量 
    int dis[501];//更新数组，作为辅助用 
    int sum = 0;//最小路径 
    int flag; 

    memset(book,0,sizeof(book));
    book[1] = 1;
    count++;
    
    for(i = 1; i<=v;i++)
    {
        dis[i] = e[1][i];
    }
    
    while(count<v)
    {
        temp = 99999;
        for(j = 1;j<=v;j++)
        {
            if(book[j]==0&&dis[j]<temp)
            {
                temp = dis[j];
                flag = j;
            }
            
        }
        book[flag] = 1;
        count++;
        sum +=dis[flag];
        
        for(j=1;j<=v;j++)
        {
            if(book[j]==0&&dis[j]>e[flag][j])
                dis[j] = e[flag][j];
        }
    }
    
    mintree = sum;
    
    return 0 ;
}



int main()
{
    
    //freopen("data.txt","r",stdin);
    mintree = 99999;//最后输出结果  
    int minl;//外界接电需要的钱 
    int num; 

    int i,j,value;//临时变量 
    scanf("%d",&num);
    while(num--)
    {
        minl = 99999;

        
        
        scanf("%d %d",&v,&l);
        
        //初始化地图 
        for(i =1;i<=v;i++)
        {
            for(j =1;j<=v;j++)
            {
                if(i==j)
                    e[i][j] = 0;
                else
                    e[i][j] = 99999;
            }
        }
        
        
        //读入数据 
        for(int k = 1;k<=l;k++)
        {
            scanf("%d %d %d",&i,&j,&value);
            e[i][j] = value;
            e[j][i] = value;
        }
         
        //找出外界接电的最小值 
        for(i =1;i<=v;i++)
        {
            scanf("%d",h+i);
            minl = minl>h[i]?h[i]:minl;
        }
        prime();
        printf("%d\n",mintree+minl);
    }
    
    
    return 0;
}