HDOJ 2682 Tree(最小生成树prim算法)

Tree

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

Total Submission(s): 1954    Accepted Submission(s): 573

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
   
   
   
   

 

题意：两个数A和B，  A或B或者A+B是素数就表示A与B能连通。A与B连通的费用是A，B，A-B的绝对值这三个数中最小的值。

先给出n个数，问将他们都连通的最小费用。

最小生成树，判断素数是要打表。

prim算法，代码如下：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#define INF 0x3f3f3f
int map[610][610],sum,prime[1000010],n;

void is_primes()//素数打表 
{
	int i,j;
	prime[0]=prime[1]=1;
	for(i=2;i*i<=1000000;++i)
	{
		if(prime[i])
		  continue;
		for(j=i*i;j<=1000000;j=i+j)
		    prime[j]=1;
	}
}

int min(int a,int b)
{
	return a<b?a:b;
}

void prim()
{
	sum=0;
	int i,j,next,min;
	int lowcost[610],visit[610];
	memset(visit,0,sizeof(visit));
	for(i=0;i<n;++i)
	   lowcost[i]=map[0][i];
	visit[0]=1;//注意从零开始的 
	for(i=1;i<n;++i)
	{
		min=INF;
		for(j=0;j<n;++j)
		{
			if(!visit[j]&&min>lowcost[j])
			{
				min=lowcost[j];
				next=j;
			}
		}
		if(min==INF)
		{
			printf("-1\n");
			return ;
		}
		sum+=min;
		visit[next]=1;
		for(j=0;j<n;++j)
		{
			if(!visit[j]&&lowcost[j]>map[next][j])
			    lowcost[j]=map[next][j];
		}
	}
	printf("%d\n",sum);
}

int main()
{
	is_primes();
	int t,a[610],i,j;
	scanf("%d",&t);
	while(t--)
	{
		memset(map,INF,sizeof(map));
		scanf("%d",&n);
		for(i=0;i<n;++i)
			scanf("%d",&a[i]);
		for(i=0;i<n;++i)
		{
			for(j=0;j<n;++j)
			{
				if(!prime[a[i]]||!prime[a[j]]||!prime[a[i]+a[j]])//满足素数条件记录入数组 
				   map[i][j]=map[j][i]=min(min(a[i],a[j]),abs(a[i]-a[j]));
			}
		}
		prim();
	}
	return 0;
}