HDOJ 2682 Tree（最小生成树--prime）

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1921    Accepted Submission(s): 561

Problem Description

There are N (2<=N<=600) cities,each has a value of happiness,we consider two cities A and B whose value of happiness are VA and VB,if VA is a prime number,or VB is a prime number or (VA+VB) is a prime number,then they can be connected.What's more,the cost to connecte two cities is Min(Min(VA , VB),|VA-VB|).
Now we want to connecte all the cities together,and make the cost minimal.

 

Input

The first will contain a integer t,followed by t cases.
Each case begin with a integer N,then N integer Vi(0<=Vi<=1000000).

 

Output

If the all cities can be connected together,output the minimal cost,otherwise output "-1";

 

Sample Input

   
   
   
   

    
    
    
    2
5
1
2
3
4
5

4
4
4
4
4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
-1




ac代码：

    
    
    
    
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 10010
#define INF 0xfffffff
#define min(a,b) (a>b?b:a)
using namespace std;
int pri[MAXN][MAXN];
int v[MAXN];
int dis[MAXN];
int n,sum;
int a[MAXN];
int b[10000002];
void fun()  
{  
    int i,j;  
    b[1]=1;  
    for(i=2;i<1000010*2;i++)  
    {  
        if(b[i]==0)
        {  
            for(j=i+i;j<1000010*2;j+=i)  
            {  
                b[j]=1;  
            }  
        }  
    }  
}  
void prime()
{
    int i,j,k;
    int min;
    memset(v,0,sizeof(v));
    for(i=1;i<=n;i++)
    dis[i]=pri[1][i];
    v[1]=1;
    sum=0;
    for(i=0;i<n-1;i++)
    {
        min=INF;
        for(j=1;j<=n;j++)
        {
            if(v[j]==0&&dis[j]<min)
            {
                min=dis[j];
                k=j;
            }
        }
        if(min==INF)
        break;
        v[k]=1;
        sum+=min;
        for(j=1;j<=n;j++)
        {
            if(v[j]==0&&dis[j]>pri[k][j])
            dis[j]=pri[k][j];
        }
    }
    if(i<n-2)
    sum=-1;
}
int main()
{
    int i,j,t;
    fun();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        pri[i][j]=pri[j][i]=INF;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(b[a[i]]==0||b[a[j]]==0||b[a[i]+a[j]]==0)
                {
                    pri[i][j]=pri[j][i]=min(min(a[i],a[j]),abs(a[i]-a[j]));
                }
                //printf("%d\n",pri[i][j]);
            }
        }
        prime();
        printf("%d\n",sum);
    }
    return 0;
}