hdoj 2682 Tree

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1861    Accepted Submission(s): 545

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

敲代码时有一个小地方没看到  结果一直RE纠结了俩小时，所以说要细心............

prime算法：

#include<stdio.h>

#include<math.h>

#define INF 0x3f3f3f

#define max 650

#include<string.h>

int lowdis[max],visit[max],map[max][max];

int city;

int sushu[1000100];

void prime()

{

	int j,i,min,mindis=0,next;

	memset(visit,0,sizeof(visit));

	for(i=1;i<=city;i++)

	{

		lowdis[i]=map[1][i];

	}

	visit[1]=1;

	for(i=2;i<=city;i++)

	{

		min=INF;

		for(j=1;j<=city;j++)

		{

			if(!visit[j]&&min>lowdis[j])

			{

				next=j;

				min=lowdis[j];

			}

		}

		if(min==INF)

		{

			printf("-1\n");

			return ;

		}

		visit[next]=1;

		mindis+=min;

		for(j=1;j<=city;j++)

		{

			if(!visit[j]&&lowdis[j]>map[next][j])

			{

				lowdis[j]=map[next][j];

			}

		}

	}

	printf("%d\n",mindis);

} 

int min(int a,int b)

{

	if(a>b)

	a=b;

	return a;

}

void biao()

{

	int i,j;

	memset(sushu,0,sizeof(sushu));

	for(i=2;i<=1000100;i++)

	{

		if(!sushu[i])

		{

			for(j=i*2;j<=1000100;j+=i)

			sushu[j]=1;

		}

	}

	sushu[1]=1;

}

int main()

{

	int n,i,j;

	int a[max];

	scanf("%d",&n);

	biao();

	while(n--)

	{

		scanf("%d",&city);

		for(i=1;i<=city;i++)

		{

			scanf("%d",&a[i]);

			for(j=1;j<=city;j++)

			{

				if(i==j)

				map[i][j]=0;

				else

				map[i][j]=map[j][i]=INF;

			}

		}

		for(i=1;i<=city;i++)

		{

			for(j=i+1;j<=city;j++)

			{

				if(!sushu[a[i]]||!sushu[a[j]]||!sushu[a[i]+a[j]])

				{

					map[i][j]=map[j][i]=min(min(a[i],a[j]),abs(a[i]-a[j]));

				}

			}

		}

		prime();

	}

	return 0;

}