HDOJ 题目2682 Tree（最小生成树）

Tree

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1757 Accepted Submission(s): 511

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

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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ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define INF 0xfffffff
#define min(a,b) (a>b?b:a)
int prime[1000010*2],a[1660],map[1010][1010],vis[1660],n;
void fun()
{
	int i,j;
	prime[1]=1;
	for(i=2;i<1000010*2;i++)
	{
		if(!prime[i])
		{
			for(j=i+i;j<1000010*2;j+=i)
			{
				prime[j]=1;
			}
		}
	}
}
int prim()
{
	int i,j;
	int sum=0;
	memset(vis,0,sizeof(vis));
	vis[1]=1;
	for(i=1;i<n;i++)
	{
		int flag=-1;
		int min=INF;
		for(j=1;j<=n;j++)
		{
			if(!vis[j]&&map[1][j]<min)
			{
				min=map[1][j];
				flag=j;
			}
			
		}
		if(flag==-1)
			break;
		sum+=map[1][flag];
		vis[flag]=1;
		for(j=1;j<=n;j++)
		{
			if(!vis[j]&&map[1][j]>map[flag][j])
			{
				map[1][j]=map[flag][j];
			}
		}
	}
	if(i<n)
		return -1;
	else
		return sum;
}
int main()
{
	int t;
	scanf("%d",&t);
	fun();
	while(t--)
	{
		//int n;
		int i,j;
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}
		//memset(map,0,sizeof(map));
		for(i=0;i<=n;i++)
		{
			for(j=0;j<=n;j++)
				map[i][j]=INF;
		}
		for(i=1;i<=n;i++)
		{
			for(j=i+1;j<=n;j++)
			{
				if(!prime[a[i]]||!prime[a[j]]||!prime[a[i]+a[j]])
				{
					map[i][j]=map[j][i]=min(min(a[i],a[j]),abs(a[i]-a[j]));
				}
			}
		}
		printf("%d\n",prim());
	}
}

HDOJ 标题2682 Tree（最小生成树）

Tree

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