hdu 2682 Tree kruskal+并查集

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2333    Accepted Submission(s): 710

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

 

没什么好说的，最小生成树 kruskal+最小生成树，模板题

 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#define MAX_N 1000005

using namespace std;

int n,m;
int par[MAX_N];
int num[MAX_N];

struct node
{
    int a,b,va;
}edge[MAX_N];
bool cmp(node a,node b)
{
    return a.va<b.va;
}
int su(int a)
{
    int flag=1;
    for(int i = 2; i <= sqrt(a)&&flag; i++)
    {
        if(a%i==0)
            flag=0;
    }
    if(a==1)
        flag=0;
    return flag;
}
int ffind(int x)
{
    if(x==par[x])
        return x;
    else
        return par[x]=ffind(par[x]);
}
bool same(int x,int y)
{
    return ffind(x)==ffind(y);
}
void mix(int x,int y)
{
    x=ffind(x);
    y=ffind(y);
    if(x!=y)
        par[x]=y;
}
int kruskal()
{
    sort(edge,edge+m,cmp);
    int ans=0,sum=0;
    for(int i = 0; i < m; i++)
    {
        if(!same(edge[i].a,edge[i].b))
        {
            mix(edge[i].a,edge[i].b);
            sum+=edge[i].va;
            ans++;
        }
    }
    if(ans==n-1)
        return sum;
    else
        return -1;

}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 0; i <= n; i++)
            par[i]=i;
        for(int i = 1; i <= n; i++)
            scanf("%d",&num[i]);
        m=0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = i+1; j <= n; j++)
            {
                if(su(num[i])||su(num[j])||su(num[i]+num[j]))
                {
                    edge[m].a=i;
                    edge[m].b=j;
                    edge[m++].va=min(min(num[i],num[j]),abs(num[i]-num[j]));
                }
            }
        }
        printf("%d
",kruskal());
    }
    return 0;
}