Tree（prime）

Tree

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

Total Submission(s) : 2   Accepted Submission(s) : 2

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

 题解：

就是给n个数，如果这个数跟另一个数的和或者这两个数有一个是素数就可以连接；权值为Min(Min(VA , VB),|VA-VB|).

注意要打素数表；

prime代码:

 #include<stdio.h>
 #include <string.h>
 #include<math.h>
 #define Min(x,y) (x<y?x:y)
 const int INF=0x3f3f3f3f;
 const int MAXN=;
 bool prim[];
 int N,answer;
 int judge(int a,int b){
         if(!prim[a]||!prim[b]||!prim[a+b]){
             return Min(Min(a,b),fabs(a-b));
         }
         else return -;
 }
 void pt(){
     memset(prim,false,sizeof(prim));
     prim[]=prim[]=true;
     for(int i=;i<=;i++){
         if(!prim[i])for(int j=i*i;j<;j+=i){
             prim[j]=true;
         }
     }
 }
 int map[MAXN][MAXN],vis[MAXN],low[MAXN];
 int v[MAXN];
 void prime(){
     int k;
     int temp,flot=;
     answer=;
     memset(vis,,sizeof(vis));
     vis[]=;
     for(int i=;i<N;i++)low[i]=map[][i];
     for(int i=;i<N;i++){
             temp=INF;
         for(int j=;j<N;j++)
             if(!vis[j]&&temp>low[j])
                     temp=low[k=j];
     if(temp==INF){
         if(flot==N)printf("%d\n",answer);
         else puts("-1");
         break;
     }
         answer+=temp;
         vis[k]=;
         flot++;
         for(int j=;j<N;j++)
             if(!vis[j]&&low[j]>map[k][j])
             low[j]=map[k][j];
     }
 }
 int main(){
         int T,t;
         scanf("%d",&T);
         while(T--){
                     pt();
                 memset(map,INF,sizeof(map));
             scanf("%d",&N);
             for(int i=;i<N;i++)scanf("%d",&v[i]);
             for(int i=;i<N;i++)
             for(int j=i+;j<N;j++){
                     t=judge(v[i],v[j]);
                 if(t>=){
                     if(t<map[i][j])map[i][j]=map[j][i]=t;
                 }
             }
             prime();
         }
     return ;
 }