【最小生成树入门专题1】H

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

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#define inf 9999999 

int e[1100][1100];
int book[1100],a[1100];
int dis[1100];
int isprime[2010000];

int Min(int a,int b)
{
	if( a<b)
		return a;
	return b;
}

int Prime(int n)
{
	int i;
	if(n ==1||n==2)
		return 0;
	for(i = 2; i *i <= n; i ++)
	{
		if(n%i == 0)
			return 0;
	}
	return 1;
}

int main()
{
	int t;
	int n,i,j,u,count,sum,min,flag;
	for(i = 1; i <= 2001000; i ++)
	{
		isprime[i] = 1;
	}
	isprime[0] = isprime[1] = 0;
	for(i = 1; i*i <= 2001000; i ++)
	{
		if(isprime[i])
		{
			for(j = i*2; j <= 2001000; j = j+i)
			{
				isprime[j] = 0;
			}
		}
	}
	scanf("%d",&t);
	while(t--)
	{
		memset(a,0,sizeof(a));
        memset(dis,0,sizeof(dis));
        memset(book,0,sizeof(book));
        memset(e,0,sizeof(e));
		sum = count = 0;
		flag = 0;
		u = 0;
		scanf("%d",&n);
		for(i = 1; i <= n; i ++)
			scanf("%d",&a[i]);
		for(i = 1; i <= n; i ++)
			for(j = 1; j <= n; j ++)
			{
				if(i == j)
					e[i][j] = 0;
				else if(isprime[a[i]]||isprime[a[j]]||isprime[a[i]+a[j]])
				{
					e[i][j] = Min(abs(a[i]-a[j]),Min(a[i],a[j]));
					e[j][i] = Min(abs(a[i]-a[j]),Min(a[i],a[j]));
					
				}
				else
				{
					e[i][j] = inf;
					e[j][i] = inf;
				}
			}
		for(i = 1; i <= n; i ++)
		{
			dis[i] = e[1][i];
		}
		count ++;
		book[1] = 1;
		while(count < n)
		{
			min = inf;
			for(i = 1; i <= n; i ++)
			{
				if(!book[i]&&dis[i]<min)
				{
					min = dis[i];
					u = i;
				}
			}
			book[u] = 1;
			sum = sum + dis[u];
			count ++;
			for(i = 1; i <= n; i ++)
			{
				if(!book[i]&&dis[i]>e[u][i])
				{
					dis[i] = e[u][i];
				}
					
			}
		}
		for(i = 1; i <= n; i ++)
			if(book[i])
				flag++;
		
		if(flag != n)
			printf("-1
");
		else
			printf("%d
",sum);
	}
	return 0;
}