Tree

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1924    Accepted Submission(s): 563

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]));
				}
			}
		}
		/*for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
		printf("%d ",map[i][j]);*/
		printf("%d
",prim());
	}
	return 0;
}