hdu 2682 Tree 最小生成树 （并查集）

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

2
5
1
2
3
4
5

4
4
4
4
4

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 : p
= find(p
);
}

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;
}