hdoj--2682--Tree（）

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2107    Accepted Submission(s): 610

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

 

Author

Teddy

 

Source

2009浙江大学计算机研考复试（机试部分）——全真模拟

 

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#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define INF 0xfffffff
#define min(a,b)(a>b?b:a)
int map[1010][1010],p[1000010*2],mark[1010],num[1010];
int n;
void fun()
{
    int i,j;
    p[1]=1;
    for(i=2;i<1000010*2;i++)
    {
        if(!p[i])
        {
            for(j=i+i;j<1000010*2;j+=i)
            {
                p[j]=1;
            }
        }
    }
}
int prim()
{
    int sum=0,p=n,i,j;
    int flog;
    memset(mark,0,sizeof(mark));
    while(--p)
    {
        int min=INF;
        for(i=2;i<=n;i++)
        {
            if(!mark[i]&&map[1][i]<min)
            {
                min=map[1][i];
                flog=i;
            }
        }
        if(min==INF)
        break;
        sum+=min;
        mark[flog]=1;
        for(j=2;j<=n;j++)
        {
            if(!mark[j]&&map[1][j]>map[flog][j])
            map[1][j]=map[flog][j];
        }
    }
    if(p) return -1;
    else
    return sum;
}
int main()
{
    int t;
    fun();
    scanf("%d",&t);
    while(t--)
    {
        int i,j;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        scanf("%d",&num[i]);
        for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        map[i][j]=INF;
        for(i=1;i<=n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                if(!p[num[i]]||!p[num[j]]||!p[num[i]+num[j]])
                {
                    map[j][i]=map[i][j]=min(min(num[i],num[j]),abs(num[i]-num[j]));
                }
            }
        }
        printf("%d\n",prim());
    }
    return 0;
}