hdu 2682 Tree 最小生成树（并查集）

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本文探讨了一种结合素数特性的最小生成树问题。在N个城市中，通过判断城市间幸福值的素数特性来确定连接可能性，并采用特定算法求解最小生成树，旨在寻找成本最低的连接方案。

Tree

Time Limit : 6000/2000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 26 Accepted Submission(s) : 10

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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

最小生成树：类似于畅通工程一类的，不过中间多加了素数的处理判断

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
struct inin
{
	int a;
	int b;
	int dis;
}boy[410000];

int p[660];
int prime[1000010];
int p2[660];

int cmp(inin a,inin b)
{
	return a.dis<b.dis;
}

int find(int n)
{
	return p[n]==n ? n : p[n] = find(p[n]);
}

int main()
{
	
	int T;
	int n;
	int k;
	int re;
	int sum;
	int i,j;
	prime[0]=1;
	prime[1]=1;
	prime[2]=0;
	for(i=2;i<1000000;i++)
	{
		if(!prime[i])
		{
			for(j=i*2;j<1000000;j+=i)
			{
				prime[j]=1;
			}
		}
	}
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		memset(p2,0,sizeof(p2));
		memset(boy,0,sizeof(boy));
		for(i=0;i<=n;i++)
		{
			p[i]=i;
		} 
		for(i=1;i<=n;i++)
		{
			scanf("%d",&p2[i]);
		}
		k=0;
		for(i=1;i<=n;i++)
		{
			for(j=i+1;j<=n;j++)
			{
				int temp=p2[i]-p2[j];
				if(temp<0)  temp=-temp;
			//	if(i!=j)
				if(!prime[p2[i]]||!prime[p2[j]]||!prime[p2[i]+p2[j]])
				{
					boy[k].a=i;
					boy[k].b=j;
					boy[k++].dis=min(p2[i],min(p2[j],temp));
				}
			}
		}
		sort(boy,boy+k,cmp);
		sum=0;	re=1;
		for(i=0;i<k;i++)
		{
			int f1=find(boy[i].a);
			int f2=find(boy[i].b);
			if(f1!=f2)
			{
				sum+=boy[i].dis;
				p[f1]=f2;
				re++;
			}
		}
		if(re!=n)
		{
			printf("-1\n");
		}
		else
		{
			printf("%d\n",sum);
		}
	}
return 0;
}