hdu2682之最小生成树

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1364    Accepted Submission(s): 401

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

251234544444

 

Sample Output

4-1

//http://acm.hdu.edu.cn/showproblem.php?pid=2682#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<cmath>#include<iomanip>#define INF 99999999using namespace std;const int MAX=600+10;int father[MAX],rank[MAX],s[MAX],sum,num;bool prime[2000009];void Prime(){prime[0]=prime[1]=true;for(int i=2;i*2<=2000000;++i)prime[i*2]=true;for(int i=3;i*i<=2000000;i+=2){if(!prime[i]){for(int j=i*i;j<=2000000;j+=2*i){prime[j]=true;}}}}struct Edge{int u,v,w;Edge(){}Edge(int U,int V,int W):u(U),v(V),w(W){}}edge[MAX*MAX];bool cmp(Edge a,Edge b){return a.w<b.w;}void makeset(int num){for(int i=0;i<=num;++i){father[i]=i;rank[i]=1;}}int findset(int v){if(v != father[v])father[v]=findset(father[v]);return father[v];}void Union(int x,int y,int l){int a=findset(x);int b=findset(y);if(a == b)return;if(rank[a]<rank[b]){father[a]=b;rank[b]+=rank[a];}else{father[b]=a;rank[a]+=rank[b];}sum+=l;--num;}int main(){Prime();int t,n;cin>>t;while(t--){cin>>n;makeset(n);int size=0;for(int i=0;i<n;++i){scanf("%d",&s[i]);for(int j=0;j<i;++j){if(!prime[s[j]] || !prime[s[i]] || !prime[s[i]+s[j]]){edge[size++]=Edge(i,j,min(min(s[i],s[j]),abs(s[i]-s[j])));}}}sum=0;num=n;sort(edge,edge+size,cmp);for(int i=0;i<size;++i)Union(edge[i].u,edge[i].v,edge[i].w);if(num == 1)cout<<sum<<endl;else cout<<"-1"<<endl;}return 0;}